Contemporary Abstract Algebra

Contemporary Abstract Algebra SEVENTH EDITION

Joseph A. Gallian University of Minnesota Duluth

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Contemporary Abstract Algebra, Seventh Edition Joseph A. Gallian VP/Editor-in-Chief: Michelle Julet Publisher: Richard Stratton Senior Sponsoring Editor: Molly Taylor Associate Editor: Daniel Seibert Editorial Assistant: Shaylin Walsh

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Contents

Preface xi PART 1

Integers and Equivalence Relations 1 0 Preliminaries 3 Properties of Integers 3 | Modular Arithmetic 7 | Mathematical Induction 12 | Equivalence Relations 15 | Functions (Mappings) 18 Exercises 21 Computer Exercises 25

PART 2

Groups 27 1 Introduction to Groups 29 Symmetries of a Square 29 | The Dihedral Groups 32 Exercises 35 Biography of Niels Abel 39

2 Groups 40 Definition and Examples of Groups 40 | Elementary Properties of Groups 48 | Historical Note 51 Exercises 52 Computer Exercises 55

3 Finite Groups; Subgroups 57 Terminology and Notation 57 | Subgroup Tests 58 | Examples of Subgroups 61 Exercises 64 Computer Exercises 70

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4 Cyclic Groups 72 Properties of Cyclic Groups 72 | Classification of Subgroups of Cyclic Groups 77 Exercises 81 Computer Exercises 86 Biography of J. J. Sylvester 89 Supplementary Exercises for Chapters 1–4 91

5 Permutation Groups 95 Definition and Notation 95 | Cycle Notation 98 | Properties of Permutations 100 | A Check Digit Scheme Based on D5 110 Exercises 113 Computer Exercises 118 Biography of Augustin Cauchy 121

6 Isomorphisms 122 Motivation 122 | Definition and Examples 122 | Cayley’s Theorem 126 | Properties of Isomorphisms 128 | Automorphisms 129 Exercises 133 Computer Exercise 136 Biography of Arthur Cayley 137

7 Cosets and Lagrange’s Theorem 138 Properties of Cosets 138 | Lagrange’s Theorem and Consequences 141 | An Application of Cosets to Permutation Groups 145 | The Rotation Group of a Cube and a Soccer Ball 146 Exercises 149 Computer Exercise 153 Biography of Joseph Lagrange 154

8 External Direct Products 155 Definition and Examples 155 | Properties of External Direct Products 156 | The Group of Units Modulo n as an External Direct Product 159 | Applications 161 Exercises 167 Computer Exercises 170 Biography of Leonard Adleman 173 Supplementary Exercises for Chapters 5–8 174

Contents

9 Normal Subgroups and Factor Groups 178 Normal Subgroups 178 | Factor Groups 180 | Applications of Factor Groups 185 | Internal Direct Products 188 Exercises 193 Biography of Évariste Galois 199

10 Group Homomorphisms 200 Definition and Examples 200 | Properties of Homomorphisms 202 | The First Isomorphism Theorem 206 Exercises 211 Computer Exercise 216 Biography of Camille Jordan 217

11 Fundamental Theorem of Finite Abelian Groups 218 The Fundamental Theorem 218 | The Isomorphism Classes of Abelian Groups 218 | Proof of the Fundamental Theorem 223 Exercises 226 Computer Exercises 228 Supplementary Exercises for Chapters 9–11 230 PART 3

Rings 235 12 Introduction to Rings 237 Motivation and Definition 237 | Examples of Rings 238 | Properties of Rings 239 | Subrings 240 Exercises 242 Computer Exercises 245 Biography of I. N. Herstein 248

13 Integral Domains 249 Definition and Examples 249 | Fields 250 | Characteristic of a Ring 252 Exercises 255 Computer Exercises 259 Biography of Nathan Jacobson 261

14 Ideals and Factor Rings 262 Ideals 262 | Factor Rings 263 | Prime Ideals and Maximal Ideals 267 Exercises 269

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Contents

Computer Exercises 273 Biography of Richard Dedekind 274 Biography of Emmy Noether 275 Supplementary Exercises for Chapters 12–14 276

15 Ring Homomorphisms 280 Definition and Examples 280 | Properties of Ring Homomorphisms 283 | The Field of Quotients 285 Exercises 287

16 Polynomial Rings 293 Notation and Terminology 293 | The Division Algorithm and Consequences 296 Exercises 300 Biography of Saunders Mac Lane 304

17 Factorization of Polynomials 305 Reducibility Tests 305 | Irreducibility Tests 308 | Unique Factorization in Z[x] 313 | Weird Dice: An Application of Unique Factorization 314 Exercises 316 Computer Exercises 319 Biography of Serge Lang 321

18 Divisibility in Integral Domains 322 Irreducibles, Primes 322 | Historical Discussion of Fermat’s Last Theorem 325 | Unique Factorization Domains 328 | Euclidean Domains 331 Exercises 335 Computer Exercise 337 Biography of Sophie Germain 339 Biography of Andrew Wiles 340 Supplementary Exercises for Chapters 15–18 341 PART 4

Fields 343 19 Vector Spaces 345 Definition and Examples 345 | Subspaces 346 | Linear Independence 347

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Exercises 349 Biography of Emil Artin 352 Biography of Olga Taussky-Todd 353

20 Extension Fields 354 The Fundamental Theorem of Field Theory 354 | Splitting Fields 356 | Zeros of an Irreducible Polynomial 362 Exercises 366 Biography of Leopold Kronecker 369

21 Algebraic Extensions 370 Characterization of Extensions 370 | Finite Extensions 372 | Properties of Algebraic Extensions 376 | Exercises 378 Biography of Irving Kaplansky 381

22 Finite Fields 382 Classification of Finite Fields 382 | Structure of Finite Fields 383 | Subfields of a Finite Field 387 Exercises 389 Computer Exercises 391 Biography of L. E. Dickson 392

23 Geometric Constructions 393 Historical Discussion of Geometric Constructions 393 | Constructible Numbers 394 | Angle-Trisectors and Circle-Squarers 396 Exercises 396 Supplementary Exercises for Chapters 19–23 399 PART 5

Special Topics 401 24 Sylow Theorems 403 Conjugacy Classes 403 | The Class Equation 404 | The Probability That Two Elements Commute 405 | The Sylow Theorems 406 | Applications of Sylow Theorems 411 Exercises 414 Computer Exercise 418 Biography of Ludwig Sylow 419

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Contents

25 Finite Simple Groups 420 Historical Background 420 | Nonsimplicity Tests 425 | The Simplicity of A5 429 | The Fields Medal 430 | The Cole Prize 430 | Exercises 431 Computer Exercises 432 Biography of Michael Aschbacher 434 Biography of Daniel Gorenstein 435 Biography of John Thompson 436

26 Generators and Relations 437 Motivation 437 | Definitions and Notation 438 | Free Group 439 | Generators and Relations 440 | Classification of Groups of Order Up to 15 444 | Characterization of Dihedral Groups 446 | Realizing the Dihedral Groups with Mirrors 447 Exercises 449 Biography of Marshall Hall, Jr. 452

27 Symmetry Groups 453 Isometries 453 | Classification of Finite Plane Symmetry Groups 455 | Classification of Finite Groups of Rotations in R3 456 Exercises 458

28 Frieze Groups and Crystallographic Groups 461 The Frieze Groups 461 | The Crystallographic Groups 467 | Identification of Plane Periodic Patterns 473 Exercises 479 Biography of M. C. Escher 484 Biography of George Pólya 485 Biography of John H. Conway 486

29 Symmetry and Counting 487 Motivation 487 | Burnside’s Theorem 488 | Applications 490 | Group Action 493 Exercises 494 Biography of William Burnside 497

30 Cayley Digraphs of Groups 498 Motivation 498 | The Cayley Digraph of a Group 498 | Hamiltonian Circuits and Paths 502 | Some Applications 508

Contents

Exercises 511 Biography of William Rowan Hamilton 516 Biography of Paul Erdös 517

31 Introduction to Algebraic Coding Theory 518 Motivation 518 | Linear Codes 523 | Parity-Check Matrix Decoding 528 | Coset Decoding 531 | Historical Note: The Ubiquitous Reed-Solomon Codes 535 Exercises 537 Biography of Richard W. Hamming 542 Biography of Jessie MacWilliams 543 Biography of Vera Pless 544

32 An Introduction to Galois Theory 545 Fundamental Theorem of Galois Theory 545 | Solvability of Polynomials by Radicals 552 | Insolvability of a Quintic 556 Exercises 557 Biography of Philip Hall 560

33 Cyclotomic Extensions 561 Motivation 561 | Cyclotomic Polynomials 562 | The Constructible Regular n-gons 566 Exercises 568 Computer Exercise 569 Biography of Carl Friedrich Gauss 570 Biography of Manjul Bhargava 571 Supplementary Exercises for Chapters 24–33 572 Selected Answers A1 Text Credits A40 Photo Credits A42 Index of Mathematicians A43 Index of Terms A45

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Preface

Dear Sir or Madam, will you read my book, it took me years to write, will you take a look? JOHN LENNON AND PAUL MCCARTNEY, Paperback Writer, single

Although I wrote the first edition of this book more than twenty years ago, my goals for it remain the same. I want students to receive a solid introduction to the traditional topics. I want readers to come away with the view that abstract algebra is a contemporary subject—that its concepts and methodologies are being used by working mathematicians, computer scientists, physicists, and chemists. I want students to enjoy reading the book. To this end, I have included lines from popular songs, poems, quotations, biographies, historical notes, dozens of photographs, hundreds of figures, numerous tables and charts, and reproductions of stamps and currency that honor mathematicians. I want students to be able to do computations and to write proofs. Accordingly, I have included an abundance of exercises to develop both skills. Changes for the seventh edition include 120 new exercises, new theorems and examples, and a freshening of the quotations and biographies. I have also expanded the supplemental material for abstract algebra available at my website. These changes accentuate and enhance the hallmark features that have made previous editions of the book a comprehensive, lively, and engaging introduction to the subject: • Extensive coverage of groups, rings, and fields, plus a variety of non-traditional special topics • A good mixture of now more than 1750 computational and theoretical exercises appearing in each chapter and in Supplementary Exercise sets that synthesize concepts from multiple chapters • Worked-out examples—now totaling 275—providing thorough practice for key concepts • Computer exercises performed using interactive software available on my website xi

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Preface

• A large number of applications from scientific and computing fields, as well as from everyday life • Numerous historical notes and biographies that illuminate the people and events behind the mathematics • Annotated suggested readings and media for interesting further exploration of topics. My website—accessible at www.d.umn.edu/~jgallian or through Cengage’s book companion site at www.cengage.com/math/gallian— offers a wealth of additional online resources supporting the book, including: • True/false questions • Flash cards • Essays on learning abstract algebra, doing proofs, and reasons why abstract algebra is a valuable subject to learn • Links to abstract algebra-related websites and software packages • . . . and much, much more. Additionally, Cengage offers the following student and instructor ancillaries to accompany the book: • A Student Solutions Manual, available for purchase separately, with worked-out solutions to the odd-numbered exercises in the book (ISBN-13: 978-0-547-16539-4; ISBN-10: 0-547-16539-0) • An online laboratory manual, written by Julianne Rainbolt and me, with exercises designed to be done with the free computer algebra system software GAP • An online Instructor’s Solutions Manual with solutions to the evennumbered exercises in the book and additional test questions and solutions • Online instructor answer keys to the book’s computer exercises and the exercises in the GAP lab manual. Connie Day was the copyeditor and Robert Messer was the accuracy reviewer. I am grateful to each of them for their careful reading of the manuscript. I also wish to express my appreciation to Janine Tangney, Daniel Seibert, and Molly Taylor from Cengage Learning, as well as Tamela Ambush and the Cengage production staff. I greatly valued the thoughtful input of the following people, who kindly served as reviewers for the seventh edition: Rebecca Berg, Bowie State University; Monte Boisen, University of Idaho; Tara Brendle, Louisiana State University; Jeff Clark, Elon University; Carl Eckberg, San Diego State University; Tom Farmer, Miami University; Yuval Flicker, Ohio State University; Ed Hinson,

Preface

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University of New Hampshire; Gizem Karaali, Pomona College; Mohan Shrikhande, Central Michigan University; Ernie Stitzinger, North Carolina State University. Over the years, many faculty and students have kindly sent me valuable comments and suggestions. They have helped to make each edition better. I owe thanks to my UMD colleague Robert McFarland for giving me numerous exercises and comments that have been included in this edition. Please send any comments and suggestions you have to me at [email protected].

Joseph A. Gallian

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PA R T

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Integers and Equivalence Relations

For online student resources, visit this textbook’s website at http://college.hmco.com/PIC/gallian7e

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0 Preliminaries

The whole of science is nothing more than a refinement of everyday thinking. ALBERT EINSTEIN, Physics and Reality

Properties of Integers Much of abstract algebra involves properties of integers and sets. In this chapter we collect the properties we need for future reference. An important property of the integers, which we will often use, is the so-called Well Ordering Principle. Since this property cannot be proved from the usual properties of arithmetic, we will take it as an axiom. Well Ordering Principle Every nonempty set of positive integers contains a smallest member.

The concept of divisibility plays a fundamental role in the theory of numbers. We say a nonzero integer t is a divisor of an integer s if there is an integer u such that s 5 tu. In this case, we write t | s (read “t divides s”). When t is not a divisor of s, we write t B s. A prime is a positive integer greater than 1 whose only positive divisors are 1 and itself. We say an integer s is a multiple of an integer t if there is an integer u such that s 5 tu. As our first application of the Well Ordering Principle, we establish a fundamental property of integers that we will use often. Theorem 0.1 Division Algorithm Let a and b be integers with b . 0. Then there exist unique integers q and r with the property that a 5 bq 1 r, where 0 # r , b.

PROOF We begin with the existence portion of the theorem. Consider the set S 5 {a 2 bk | k is an integer and a 2 bk $ 0}. If 0 [ S, then b 3

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Integers and Equivalence Relations

divides a and we may obtain the desired result with q 5 a/b and r 5 0. Now assume 0 n S. Since S is nonempty [if a . 0, a 2 b ? 0 [ S; if a , 0, a 2 b(2a) 5 a(1 2 2b) [ S; a  0 since 0 n S], we may apply the Well Ordering Principle to conclude that S has a smallest member, say r 5 a 2 bq. Then a 5 bq 1 r and r $ 0, so all that remains to be proved is that r , b. If r $ b, then a 2 b(q 1 1) 5 a 2 bq 2 b 5 r 2 b $ 0, so that a 2 b(q 1 1) [ S. But a 2 b(q 1 1) , a 2 bq, and a 2 bq is the smallest member of S. So, r , b. To establish the uniqueness of q and r, let us suppose that there are integers q, q9, r, and r9 such that a 5 bq 1 r,

0#r,b

and

a 5 bq9 1 r9,

0 # r9 , b.

For convenience, we may also suppose that r9 $ r. Then bq 1 r 5 bq9 1 r9 and b(q 2 q9) 5 r9 2 r. So, b divides r9 2 r and 0 # r9 2 r # r9 , b. It follows that r9 2 r 5 0, and therefore r9 5 r and q 5 q9. The integer q in the division algorithm is called the quotient upon dividing a by b; the integer r is called the remainder upon dividing a by b. EXAMPLE 1 For a 5 17 and b 5 5, the division algorithm gives 17 5 5 ? 3 1 2; for a 5 223 and b 5 6, the division algorithm gives 223 5 6(24) 1 1. Several states use linear functions to encode the month and date of birth into a three-digit number that is incorporated into driver’s license numbers. If the encoding function is known, the division algorithm can be used to recapture the month and date of birth from the three-digit number. For instance, the last three digits of a Florida male driver’s license number are those given by the formula 40(m 2 1) 1 b, where m is the number of the month of birth and b is the day of birth. Thus, since 177 5 40 ? 4 1 17, a person with these last three digits was born on May 17. For New York licenses issued prior to September of 1992, the last two digits indicate the year of birth, and the three preceding digits code the month and date of birth. For a male driver, these three digits are 63m 1 2b, where m denotes the number of the month of birth and b is the date of birth. So, since 701 5 63 ? 11 1 2 ? 4, a license that ends with 70174 indicates that the holder is a male born on November 4, 1974. (In cases where the formula for the driver’s license number yields the same result for two or more people, a “tie-breaking” digit is inserted before the two digits

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5

for the year of birth.) Incidentally, Wisconsin uses the same method as Florida to encode birth information, but the numbers immediately precede the last pair of digits. Definitions Greatest Common Divisor, Relatively Prime Integers

The greatest common divisor of two nonzero integers a and b is the largest of all common divisors of a and b. We denote this integer by gcd(a, b). When gcd(a, b) 5 1, we say a and b are relatively prime.

The following property of the greatest common divisor of two integers plays a critical role in abstract algebra. The proof provides an application of the division algorithm and our second application of the Well Ordering Principle. Theorem 0.2 GCD Is a Linear Combination For any nonzero integers a and b, there exist integers s and t such that gcd(a, b) 5 as 1 bt. Moreover, gcd(a, b) is the smallest positive integer of the form as 1 bt.

PROOF Consider the set S 5 {am 1 bn | m, n are integers and am 1 bn . 0}. Since S is obviously nonempty (if some choice of m and n makes am 1 bn , 0, then replace m and n by 2m and 2n), the Well Ordering Principle asserts that S has a smallest member, say, d 5 as 1 bt. We claim that d 5 gcd(a, b). To verify this claim, use the division algorithm to write a 5 dq 1 r, where 0 # r , d. If r . 0, then r 5 a 2 dq 5 a 2 (as 1 bt)q 5 a 2 asq 2 btq 5 a(1 2 sq) 1 b(2tq) [ S, contradicting the fact that d is the smallest member of S. So, r 5 0 and d divides a. Analogously (or, better yet, by symmetry), d divides b as well. This proves that d is a common divisor of a and b. Now suppose d9 is another common divisor of a and b and write a 5 d9h and b 5 d9k. Then d 5 as 1 bt 5 (d9h)s 1 (d9k)t 5 d9(hs 1 kt), so that d9 is a divisor of d. Thus, among all common divisors of a and b, d is the greatest. The special case of Theorem 0.2 when a and b are relatively prime is so important in abstract algebra that we single it out as a corollary. Corollary If a and b are relatively prime, than there exist integers s and t such that as 1 bt 5 1.

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Integers and Equivalence Relations

EXAMPLE 2 gcd(4, 15) 5 1; gcd(4, 10) 5 2; gcd(22 ? 32 ? 5, 2 ? 33 ? 5 2 ? 32. Note that 4 and 15 are relatively prime, whereas 4 and 10 are not. Also, 4 ? 4 1 15(21) 5 1 and 4(22) 1 10 ? 1 5 2. 72)

The next lemma is frequently used. It appeared in Euclid’s Elements. Euclid’s Lemma p | ab Implies p | a or p | b If p is a prime that divides ab, then p divides a or p divides b.

PROOF Suppose p is a prime that divides ab but does not divide a. We must show that p divides b. Since p does not divide a, there are integers s and t such that 1 5 as 1 pt. Then b 5 abs 1 ptb, and since p divides the right-hand side of this equation, p also divides b. Note that Euclid’s Lemma may fail when p is not a prime, since 6 | (4 ? 3) but 6 B 4 and 6 B 3. Our next property shows that the primes are the building blocks for all integers. We will often use this property without explicitly saying so. Theorem 0.3 Fundamental Theorem of Arithmetic Every integer greater than 1 is a prime or a product of primes. This product is unique, except for the order in which the factors appear. That is, if n 5 p1 p2 . . . pr and n 5 q1q2 . . . qs , where the p’s and q’s are primes, then r 5 s and, after renumbering the q’s, we have pi 5 qi for all i.

We will prove the existence portion of Theorem 0.3 later in this chapter. The uniqueness portion is a consequence of Euclid’s Lemma (Exercise 27). Another concept that frequently arises is that of the least common multiple of two integers. Definition Least Common Multiple

The least common multiple of two nonzero integers a and b is the smallest positive integer that is a multiple of both a and b. We will denote this integer by lcm(a, b).

We leave it as an exercise (Exercise 12) to prove that every common multiple of a and b is a multiple of lcm(a, b).

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EXAMPLE 3 lcm(4, 6) 5 12; lcm(4, 8) 5 8; lcm(10, 12) 5 60; lcm(6, 5) 5 30; lcm(22 ? 32 ? 5, 2 ? 33 ? 72) 5 22 ? 33 ? 5 ? 72.

Modular Arithmetic Another application of the division algorithm that will be important to us is modular arithmetic. Modular arithmetic is an abstraction of a method of counting that you often use. For example, if it is now September, what month will it be 25 months from now? Of course, the answer is October, but the interesting fact is that you didn’t arrive at the answer by starting with September and counting off 25 months. Instead, without even thinking about it, you simply observed that 25 5 2 ? 12 1 1, and you added 1 month to September. Similarly, if it is now Wednesday, you know that in 23 days it will be Friday. This time, you arrived at your answer by noting that 23 5 7 ? 3 1 2, so you added 2 days to Wednesday instead of counting off 23 days. If your electricity is off for 26 hours, you must advance your clock 2 hours, since 26 5 2 ? 12 1 2. Surprisingly, this simple idea has numerous important applications in mathematics and computer science. You will see a few of them in this section. The following notation is convenient. When a 5 qn 1 r, where q is the quotient and r is the remainder upon dividing a by n, we write a mod n 5 r. Thus, 3 mod 2 5 1 since 3 5 1 ? 2 1 1, 6 mod 2 5 0 since 6 5 3 ? 2 1 0, 11 mod 3 5 2 since 11 5 3 ? 3 1 2, 62 mod 85 5 62 since 62 5 0 ? 85 1 62, 22 mod 15 5 13 since 22 5 (21)15 1 13. In general, if a and b are integers and n is a positive integer, then a mod n 5 b mod n if and only if n divides a 2 b (Exercise 9). In our applications, we will use addition and multiplication mod n. When you wish to compute ab mod n or (a 1 b) mod n, and a or b is greater than n, it is easier to “mod first.” For example, to compute (27 ? 36) mod 11, we note that 27 mod 11 5 5 and 36 mod 11 5 3, so (27 ? 36) mod 11 5 (5 ? 3) mod 11 5 4. (See Exercise 11.) Modular arithmetic is often used in assigning an extra digit to identification numbers for the purpose of detecting forgery or errors. We present two such applications. EXAMPLE 4 The United States Postal Service money order shown in Figure 0.1 has an identification number consisting of 10 digits together with an extra digit called a check. The check digit is the 10-digit number modulo 9. Thus, the number 3953988164 has the check digit 2, since

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Integers and Equivalence Relations

Figure 0.1

3953988164 mod 9 5 2.† If the number 39539881642 were incorrectly entered into a computer (programmed to calculate the check digit) as, say, 39559881642 (an error in the fourth position), the machine would calculate the check digit as 4, whereas the entered check digit would be 2. Thus the error would be detected. EXAMPLE 5 Airline companies, United Parcel Service, and the rental car companies Avis and National use the modulo 7 values of identification numbers to assign check digits. Thus, the identification number 00121373147367 (see Figure 0.2) has the check digit 3 appended

Figure 0.2

†The

value of N mod 9 is easy to compute with a calculator. If N 5 9q 1 r, where r is the remainder upon dividing N by 9, then on a calculator screen N 4 9 appears as q.rrrrr . . . , so the first decimal digit is the check digit. For example, 3953988164 4 9 5 439332018.222, so 2 is the check digit. If N has too many digits for your calculator, replace N by the sum of its digits and divide that number by 9. Thus, 3953988164 mod 9 5 56 mod 9 5 2. The value of 3953988164 mod 9 can also be computed by searching Google for 3953988164 mod 9.

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9

Figure 0.3

to it because 121373147367 mod 7 5 3. Similarly, the UPS pickup record number 768113999, shown in Figure 0.3, has the check digit 2 appended to it. The methods used by the Postal Service and the airline companies do not detect all single-digit errors (see Exercises 35 and 39). However, detection of all single-digit errors, as well as nearly all errors involving the transposition of two adjacent digits, is easily achieved. One method that does this is the one used to assign the so-called Universal Product Code (UPC) to most retail items (see Figure 0.4). A UPC identification number has 12 digits. The first six digits identify the manufacturer, the next five identify the product, and the last is a check. (For many items, the 12th digit is not printed, but it is always bar-coded.) In Figure 0.4, the check digit is 8.

Figure 0.4

To explain how the check digit is calculated, it is convenient to introduce the dot product notation for two k-tuples: (a1, a2, . . . , ak) ? (w1, w2, . . . , wk) 5 a1w1 1 a2w2 1 ? ? ? 1 akwk.

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Integers and Equivalence Relations

An item with the UPC identification number a1a2 ??? a12 satisfies the condition (a1, a2, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0. To verify that the number in Figure 0.4 satisfies the condition above, we calculate (0 ? 3 1 2 ? 1 1 1 ? 3 1 0 ? 1 1 0 ? 3 1 0 ? 1 1 6 ? 3 1 5 ? 1 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1) mod 10 5 90 mod 10 5 0. The fixed k-tuple used in the calculation of check digits is called the weighting vector. Now suppose a single error is made in entering the number in Figure 0.4 into a computer. Say, for instance, that 021000958978 is entered (notice that the seventh digit is incorrect). Then the computer calculates 0?312?111?310?110?310?119?3 1 5 ? 1 1 8 ? 3 1 9 ? 1 1 7 ? 3 1 8 ? 1 5 99. Since 99 mod 10  0, the entered number cannot be correct. In general, any single error will result in a sum that is not 0 modulo 10. The advantage of the UPC scheme is that it will detect nearly all errors involving the transposition of two adjacent digits as well as all errors involving one digit. For doubters, let us say that the identification number given in Figure 0.4 is entered as 021000658798. Notice that the last two digits preceding the check digit have been transposed. But by calculating the dot product, we obtain 94 mod 10  0, so we have detected an error. In fact, the only undetected transposition errors of adjacent digits a and b are those where |a 2 b| 5 5. To verify this, we observe that a transposition error of the form a1a2 ? ? ? aiai11 ? ? ? a12 → a1a2 ? ? ? ai11ai ? ? ? a12 is undetected if and only if (a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 0. That is, the error is undetected if and only if (a1, a2, . . . , ai11, ai, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10 5 (a1, a2, . . . , ai, ai11, . . . , a12) ? (3, 1, 3, 1, . . . , 3, 1) mod 10. This equality simplifies to either (3ai11 1 ai) mod 10 5 (3ai 1 ai11) mod 10 or (ai11 1 3ai) mod 10 5 (ai 1 3ai11) mod 10

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depending on whether i is even or odd. Both cases reduce to 2(ai11 2 ai) mod 10 5 0. It follows that |ai11 2 ai| 5 5, if ai11  ai. In 2005 United States companies began to phase in the use of a 13th digit to be in conformance with the 13-digit product indentification numbers used in Europe. The weighing vector for 13-digit numbers is (1, 3, 1, 3, . . . , 3, 1). Identification numbers printed on bank checks (on the bottom left between the two colons) consist of an eight-digit number a1a2 ? ? ? a8 and a check digit a9, so that (a1, a2, . . . , a9) ? (7, 3, 9, 7, 3, 9, 7, 3, 9) mod 10 5 0. As is the case for the UPC scheme, this method detects all singledigit errors and all errors involving the transposition of adjacent digits a and b except when |a 2 b| 5 5. But it also detects most errors of the form ? ? ? abc ? ? ? → ? ? ? cba ? ? ?, whereas the UPC method detects no errors of this form. In Chapter 5, we will examine more sophisticated means of assigning check digits to numbers. What about error correction? Suppose you have a number such as 73245018 and you would like to be sure that even if a single mistake were made in entering this number into a computer, the computer would nevertheless be able to determine the correct number. (Think of it. You could make a mistake in dialing a telephone number but still get the correct phone to ring!) This is possible using two check digits. One of the check digits determines the magnitude of any single-digit error, while the other check digit locates the position of the error. With these two pieces of information, you can fix the error. To illustrate the idea, let us say that we have the eight-digit identification number a1a2 ? ? ? a8. We assign two check digits a9 and a10 so that (a1 1 a2 1 ? ? ? 1 a9 1 a10) mod 11 5 0 and (a1, a2, . . . , a9, a10) ? (1, 2, 3, . . . , 10) mod 11 5 0 are satisfied. Let’s do an example. Say our number before appending the two check digits is 73245018. Then a9 and a10 are chosen to satisfy (7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 1 a9 1 a10) mod 11 5 0

(1)

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Integers and Equivalence Relations

and (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1 0 ? 6 1 1 ? 7 1 8 ? 8 1 a9 ? 9 1 a10 ? 10) mod 11 5 0.

(2)

Since 7 1 3 1 2 1 4 1 5 1 0 1 1 1 8 5 30 and 30 mod 11 5 8, Equation (1) reduces to (8 1 a9 1 a10 ) mod 11 5 0.

(19)

Likewise, since (7 ? 1 1 3 ? 2 1 2 ? 3 1 4 ? 4 1 5 ? 5 1 0 ? 6 1 1 ? 7 1 8 ? 8) mod 11 5 10, Equation (2) reduces to (10 1 9a9 1 10a10) mod 11 5 0.

(29)

Since we are using mod 11, we may rewrite Equation (29) as (21 2 2a9 2 a10) mod 11 5 0 and add this to Equation (19) to obtain 7 2 a9 5 0. Thus a9 5 7. Now substituting a9 5 7 into Equation (19) or Equation (29), we obtain a10 5 7 as well. So, the number is encoded as 7324501877. Now let us suppose that this number is erroneously entered into a computer programmed with our encoding scheme as 7824501877 (an error in position 2). Since the sum of the digits of the received number mod 11 is 5, we know that some digit is 5 too large or 6 too small (assuming only one error has been made). But which one? Say the error is in position i. Then the second dot product has the form a1 ? 1 1 a2 ? 2 1 ? ? ? 1 (ai 1 5)i 1 ai11 ? (i 1 1) 1 ? ? ? 1 a10 ? 10 5 (a1, a2, ? ? ? , a10) ? (1, 2, ? ? ? , 10) 1 5i. So, (7, 8, 2, 4, 5, 0, 1, 8, 7, 7) ? (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) mod 11 5 5i mod 11. Since the left-hand side mod 11 is 10, we see that i 5 2. Our conclusion: The digit in position 2 is 5 too large. We have successfully corrected the error.

Mathematical Induction There are two forms of proof by mathematical induction that we will use. Both are equivalent to the Well Ordering Principle. The explicit formulation of the method of mathematical induction came in the 16th century. Francisco Maurolycus (1494–1575), a teacher of Galileo, used it in 1575 to prove that 1 1 3 1 5 1 ? ? ? 1 (2n 2 1) 5 n2, and Blaise Pascal (1623–1662) used it when he presented what we now call Pascal’s triangle for the coefficients of the binomial expansion. The term mathematical induction was coined by Augustus De Morgan.

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Theorem 0.4 First Principle of Mathematical Induction Let S be a set of integers containing a. Suppose S has the property that whenever some integer n $ a belongs to S, then the integer n 1 1 also belongs to S. Then, S contains every integer greater than or equal to a.

PROOF The proof is left as an exercise (Exercise 29). So, to use induction to prove that a statement involving positive integers is true for every positive integer, we must first verify that the statement is true for the integer 1. We then assume the statement is true for the integer n and use this assumption to prove that the statement is true for the integer n 1 1. Our next example uses some facts about plane geometry. Recall that given a straightedge and compass, we can construct a right angle. EXAMPLE 6 We use induction to prove that given a straightedge, a compass, and a unit length, we can construct a line segment of length "n for every positive integer n. The case when n 5 1 is given. Now we assume that we can construct a line segment of length "n . Then use the straightedge and compass to construct a right triangle with height 1 and base "n . The hypotenuse of the triangle has length "n 1 1 . So, by induction, we can construct a line segment of length "n for every positive integer n. EXAMPLE 7 DEMOIVRE’S THEOREM We use induction to prove that for every positive integer n and every real number u, (cos u 1 i sin u)n 5 cos nu 1 i sin nu, where i is the complex number " 21 . Obviously, the statement is true for n 5 1. Now assume it is true for n. We must prove that (cos u 1 i sin u)n11 5 cos(n 1 1)u 1 i sin(n 1 1)u. Observe that (cos u 1 i sin u)n11 5 (cos u 1 i sin u)n(cos u 1 i sin u) 5 (cos nu 1 i sin nu)(cos u 1 i sin u) 5 cos nu cos u 1 i(sin nu cos u 1 sin u cos nu) 2 sin nu sin u. Now, using trigonometric identities for cos(a 1 b) and sin(a 1 b), we see that this last term is cos(n 1 1)u 1 i sin(n 1 1)u. So, by induction, the statement is true for all positive integers. In many instances, the assumption that a statement is true for an integer n does not readily lend itself to a proof that the statement is true

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for the integer n 1 1. In such cases, the following equivalent form of induction may be more convenient. Some authors call this formulation the strong form of induction. Theorem 0.5 Second Principle of Mathematical Induction Let S be a set of integers containing a. Suppose S has the property that n belongs to S whenever every integer less than n and greater than or equal to a belongs to S. Then, S contains every integer greater than or equal to a.

PROOF The proof is left to the reader. To use this form of induction, we first show that the statement is true for the integer a. We then assume that the statement is true for all integers that are greater than or equal to a and less than n, and use this assumption to prove that the statement is true for n. EXAMPLE 8 We will use the Second Principle of Mathematical Induction with a 5 2 to prove the existence portion of the Fundamental Theorem of Arithmetic. Let S be the set of integers greater than 1 that are primes or products of primes. Clearly, 2 [ S. Now we assume that for some integer n, S contains all integers k with 2 # k , n. We must show that n [ S. If n is a prime, then n [ S by definition. If n is not a prime, then n can be written in the form ab, where 1 , a , n and 1 , b , n. Since we are assuming that both a and b belong to S, we know that each of them is a prime or a product of primes. Thus, n is also a product of primes. This completes the proof. Notice that it is more natural to prove the Fundamental Theorem of Arithmetic with the Second Principle of Mathematical Induction than with the First Principle. Knowing that a particular integer factors as a product of primes does not tell you anything about factoring the next larger integer. (Does knowing that 5280 is a product of primes help you to factor 5281 as a product of primes?) The following problem appeared in the “Brain Boggler” section of the January 1988 issue of the science magazine Discover. EXAMPLE 9 The Quakertown Poker Club plays with blue chips worth $5.00 and red chips worth $8.00. What is the largest bet that cannot be made?

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To gain insight into this problem, we try various combinations of blue and red chips and obtain 5, 8, 10, 13, 15, 16, 18, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40. It appears that the answer is 27. But how can we be sure? Well, we need only prove that every integer greater than 27 can be written in the form a ? 5 1 b ? 8, where a and b are nonnegative integers. This will solve the problem, since a represents the number of blue chips and b the number of red chips needed to make a bet of a ? 5 1 b ? 8. For the purpose of contrast, we will give two proofs—one using the First Principle of Mathematical Induction and one using the Second Principle. Let S be the set of all integers greater than or equal to 28 of the form a ? 5 1 b ? 8, where a and b are nonnegative. Obviously, 28 [ S. Now assume that some integer n [ S, say, n 5 a ? 5 1 b ? 8. We must show that n 1 1 [ S. First, note that since n $ 28, we cannot have both a and b less than 3. If a $ 3, then n 1 1 5 (a ? 5 1 b ? 8) 1 (23 ? 5 1 2 ? 8) 5 (a 2 3) ? 5 1 (b 1 2) ? 8. (Regarding chips, this last equation says that we may increase a bet from n to n 1 1 by removing three blue chips from the pot and adding two red chips.) If b $ 3, then n 1 1 5 (a ? 5 1 b ? 8) 1 (5 ? 5 2 3 ? 8) 5 (a 1 5) ? 5 1 (b 2 3) ? 8. (The bet can be increased by 1 by removing three red chips and adding five blue chips.) This completes the proof. To prove the same statement by the Second Principle, we note that each of the integers 28, 29, 30, 31, and 32 is in S. Now assume that for some integer n . 32, S contains all integers k with 28 # k , n. We must show that n [ S. Since n 2 5 [ S, there are nonnegative integers a and b such that n 2 5 5 a ? 5 1 b ? 8. But then n 5 (a 1 1) ? 5 1 b ? 8. Thus n is in S.

Equivalence Relations In mathematics, things that are considered different in one context may be viewed as equivalent in another context. We have already seen one such example. Indeed, the sums 2 1 1 and 4 1 4 are certainly different in ordinary arithmetic, but are the same under modulo 5 arithmetic. Congruent triangles that are situated differently in the plane are not the same, but they are often considered to be the same in plane geometry. In physics, vectors of the same magnitude and direction can produce

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Integers and Equivalence Relations

different effects—a 10-pound weight placed 2 feet from a fulcrum produces a different effect than a 10-pound weight placed 1 foot from a fulcrum. But in linear algebra, vectors of the same magnitude and direction are considered to be the same. What is needed to make these distinctions precise is an appropriate generalization of the notion of equality; that is, we need a formal mechanism for specifying whether or not two quantities are the same in a given setting. This mechanism is an equivalence relation. Definition Equivalence Relation

An equivalence relation on a set S is a set R of ordered pairs of elements of S such that 1. (a, a) [ R for all a [ S (reflexive property). 2. (a, b) [ R implies (b, a) [ R (symmetric property). 3. (a, b) [ R and (b, c) [ R imply (a, c) [ R (transitive property).

When R is an equivalence relation on a set S, it is customary to write aRb instead of (a, b) [ R. Also, since an equivalence relation is just a generalization of equality, a suggestive symbol such as <, ;, or , is usually used to denote the relation. Using this notation, the three conditions for an equivalence relation become a , a; a , b implies b , a; and a , b and b , c imply a , c. If , is an equivalence relation on a set S and a [ S, then the set [a] 5 {x [ S | x , a} is called the equivalence class of S containing a. EXAMPLE 10 Let S be the set of all triangles in a plane. If a, b [ S, define a , b if a and b are similar—that is, if a and b have corresponding angles that are the same. Then, , is an equivalence relation on S. EXAMPLE 11 Let S be the set of all polynomials with real coefficients. If f, g [ S, define f , g if f 9 5 g9, where f 9 is the derivative of f. Then, , is an equivalence relation on S. Since two polynomials with equal derivatives differ by a constant, we see that for any f in S, [ f ] 5 { f 1 c | c is real}. EXAMPLE 12 Let S be the set of integers and let n be a positive integer. If a, b [ S, define a ; b if a mod n 5 b mod n (that is, if a 2 b is divisible by n). Then, ; is an equivalence relation on S and [a] 5 {a 1 kn | k [ S}. Since this particular relation is important in abstract algebra, we will take the trouble to verify that it is indeed an equivalence relation. Certainly, a 2 a is divisible by n, so that a ; a for all a in S. Next, assume that a ; b, say, a 2 b 5 rn. Then, b 2 a 5 (2r)n, and

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therefore b ; a. Finally, assume that a ; b and b ; c, say, a 2 b 5 rn and b 2 c 5 sn. Then, we have a 2 c 5 (a 2 b) 1 (b 2 c) 5 rn 1 sn 5 (r 1 s)n, so that a ; c. EXAMPLE 13 Let ; be as in Example 12 and let n 5 7. Then we have 16 ; 2; 9 ; 25; and 24 ; 3. Also, [1] 5 {. . . , 220, 213, 26, 1, 8, 15, . . .} and [4] 5 {. . . , 217, 210, 23, 4, 11, 18, . . .}. EXAMPLE 14 Let S 5 {(a, b) | a, b are integers, b 2 0}. If (a, b), (c, d ) [ S, define (a, b) < (c, d ) if ad 5 bc. Then < is an equivalence relation on S. [The motivation for this example comes from fractions. In fact, the pairs (a, b) and (c, d) are equivalent if the fractions a/b and c/d are equal.] To verify that < is an equivalence relation on S, note that (a, b) < (a, b) requires that ab 5 ba, which is true. Next, we assume that (a, b) < (c, d), so that ad 5 bc. We have (c, d) < (a, b) provided that cb 5 da, which is true from commutativity of multiplication. Finally, we assume that (a, b) < (c, d ) and (c, d) < (e, f ) and prove that (a, b) < (e, f ). This amounts to using ad 5 bc and cf 5 de to show that af 5 be. Multiplying both sides of ad 5 bc by f and replacing cf by de, we obtain adf 5 bcf 5 bde. Since d 2 0, we can cancel d from the first and last terms. Definition Partition

A partition of a set S is a collection of nonempty disjoint subsets of S whose union is S. Figure 0.5 illustrates a partition of a set into four subsets. S

Figure 0.5 Partition of S into four subsets.

EXAMPLE 15 The sets {0}, {1, 2, 3, . . .}, and {. . . , 23, 22, 21} constitute a partition of the set of integers. EXAMPLE 16 The set of nonnegative integers and the set of nonpositive integers do not partition the integers, since both contain 0. The next theorem reveals that equivalence relations and partitions are intimately intertwined.

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Theorem 0.6 Equivalence Classes Partition The equivalence classes of an equivalence relation on a set S constitute a partition of S. Conversely, for any partition P of S, there is an equivalence relation on S whose equivalence classes are the elements of P.

PROOF Let , be an equivalence relation on a set S. For any a [ S, the reflexive property shows that a [ [a]. So, [a] is nonempty and the union of all equivalence classes is S. Now, suppose that [a] and [b] are distinct equivalence classes. We must show that [a] > [b] 5 0/. On the contrary, assume c [ [a] > [b]. We will show that [a] # [b]. To this end, let x [ [a]. We then have c , a, c , b, and x , a. By the symmetric property, we also have a , c. Thus, by transitivity, x , c, and by transitivity again, x , b. This proves [a] # [b]. Analogously, [b] # [a]. Thus, [a] 5 [b], in contradiction to our assumption that [a] and [b] are distinct equivalence classes. To prove the converse, let P be a collection of nonempty disjoint subsets of S whose union is S. Define a , b if a and b belong to the same subset in the collection. We leave it to the reader to show that , is an equivalence relation on S (Exercise 55).

Functions (Mappings) Although the concept of a function plays a central role in nearly every branch of mathematics, the terminology and notation associated with functions vary quite a bit. In this section, we establish ours. Definition Function (Mapping)

A function (or mapping) f from a set A to a set B is a rule that assigns to each element a of A exactly one element b of B. The set A is called the domain of f, and B is called the range of f. If f assigns b to a, then b is called the image of a under f. The subset of B comprising all the images of elements of A is called the image of A under f.

We use the shorthand f: A → B to mean that f is a mapping from A to B. We will write f(a) 5 b or f: a → b to indicate that f carries a to b. There are often different ways to denote the same element of a set. In defining a function in such cases one must verify that the function values assigned to the elements depend not on the way the elements are expressed but only on the elements themselves. For example, the

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correspondence f from the rational numbers to the integers given by f(a/b) 5 a 1 b does not define a function since 1/2 5 2/4 but f (1/2) ? f (2/4). To verify that a correspondence is a function, you assume that x1 5 x2 and prove that f (x1) 5 (x2). Definition Composition of Functions

Let f: A → B and c: B → C. The composition cf is the mapping from A to C defined by (cf)(a) 5 c(f(a)) for all a in A. The composition function cf can be visualized as in Figure 0.6.

φ

ψ φ (a) ψ (φ (a))

a

ψφ

Figure 0.6 Composition of functions f and c.

In calculus courses, the composition of f with g is written ( f 8 g)(x) and is defined by ( f 8 g)(x) 5 f (g(x)). When we compose functions, we omit the “circle.” There are several kinds of functions that occur often enough to be given names. Definition One-to-One Function

A function f from a set A is called one-to-one if for every a1, a2 [ A, f(a1) 5 f(a2) implies a1 5 a2.

The term one-to-one is suggestive, since the definition ensures that one element of B can be the image of only one element of A. Alternatively, f is one-to-one if a1  a2 implies f(a1)  f(a2). That is, different elements of A map to different elements of B. See Figure 0.7. φ

ψ

a1

φ (a1)

a1

a2

φ (a2)

a2

ψ (a1) 5 ψ (a2)

ψ is not one-to-one

φ is one-to-one

Figure 0.7

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Definition Function from A onto B

A function f from a set A to a set B is said to be onto B if each element of B is the image of at least one element of A. In symbols, f: A → B is onto if for each b in B there is at least one a in A such that f(a) 5 b. See Figure 0.8. φ

ψ

φ is onto

ψ is not onto

Figure 0.8

The next theorem summarizes the facts about functions we will need. Theorem 0.7 Properties of Functions Given functions a: A → B, b: B → C, and g: C → D, then 1. g(ba) 5 (gb)a (associativity). 2. If a and b are one-to-one, then ba is one-to-one. 3. If a and b are onto, then ba is onto. 4. If a is one-to-one and onto, then there is a function a21 from B onto A such that (a21a)(a) 5 a for all a in A and (aa21)(b) 5 b for all b in B.

PROOF We prove only part 1. The remaining parts are left as exercises (Exercise 51). Let a [ A. Then (g(ba))(a) 5 g((ba)(a)) 5 g(b(a(a))). On the other hand, ((gb)a)(a) 5 (gb)(a(a)) 5 g(b(a(a))). So, g(ba) 5 (gb)a. It is useful to note that if a is one-to-one and onto, the function a21 described in part 4 of Theorem 0.7 has the property that if a(s) 5 t, then a21(t) 5 s. That is, the image of t under a21 is the unique element s that maps to t under a. In effect, a21 “undoes” what a does. EXAMPLE 17 Let Z denote the set of integers, R the set of real numbers, and N the set of nonnegative integers. The following table illustrates the properties of one-to-one and onto.

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Domain Z R Z Z

Range Z R N Z

Rule x → x3 x → x3 x → |x| x → x2

One-to-one Yes Yes No No

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Onto No Yes Yes No

To verify that x → x3 is one-to-one in the first two cases, notice that if x3 5 y3, we may take the cube roots of both sides of the equation to obtain x 5 y. Clearly, the mapping from Z to Z given by x → x3 is not onto, since 2 is the cube of no integer. However, x → x3 defines an onto function from R to R, since every real number is the cube of its 3 cube root (that is, " b → b). The remaining verifications are left to the reader.

Exercises I was interviewed in the Israeli Radio for five minutes and I said that more than 2000 years ago, Euclid proved that there are infinitely many primes. Immediately the host interrupted me and asked: “Are there still infinitely many primes?” NOGA ALON

1. For n 5 5, 8, 12, 20, and 25, find all positive integers less than n and relatively prime to n. 2. Determine gcd(24 ? 32 ? 5 ? 72, 2 ? 33 ? 7 ? 11) and lcm(23 ? 32 ? 5, 2 ? 33 ? 7 ? 11). 3. Determine 51 mod 13, 342 mod 85, 62 mod 15, 10 mod 15, (82 ? 73) mod 7, (51 1 68) mod 7, (35 ? 24) mod 11, and (47 1 68) mod 11. 4. Find integers s and t such that 1 5 7 ? s 1 11 ? t. Show that s and t are not unique. 5. In Florida, the fourth and fifth digits from the end of a driver’s license number give the year of birth. The last three digits for a male with birth month m and birth date b are represented by 40(m 2 1) 1 b. For females the digits are 40(m 2 1) 1 b 1 500. Determine the dates of birth of people who have last five digits 42218 and 53953. 6. For driver’s license numbers issued in New York prior to September of 1992, the three digits preceding the last two of the number of a male with birth month m and birth date b are represented by 63m 1 2b. For females the digits are 63m 1 2b 1 1. Determine the dates of birth and sex(es) corresponding to the numbers 248 and 601.

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Integers and Equivalence Relations

7. Show that if a and b are positive integers, then ab 5 lcm(a, b) ? gcd(a, b). 8. Suppose a and b are integers that divide the integer c. If a and b are relatively prime, show that ab divides c. Show, by example, that if a and b are not relatively prime, then ab need not divide c. 9. If a and b are integers and n is a positive integer, prove that a mod n 5 b mod n if and only if n divides a 2 b. 10. Let a and b be integers and d 5 gcd(a, b). If a 5 da9 and b 5 db9, show that gcd(a9, b9) 5 1. 11. Let n be a fixed positive integer greater than 1. If a mod n 5 a9 and b mod n 5 b9, prove that (a 1 b) mod n 5 (a9 1 b9) mod n and (ab) mod n 5 (a9b9) mod n. (This exercise is referred to in Chapters 6, 8, and 15.) 12. Let a and b be positive integers and let d 5 gcd(a, b) and m 5 lcm(a, b). If t divides both a and b, prove that t divides d. If s is a multiple of both a and b, prove that s is a multiple of m. 13. Let n and a be positive integers and let d 5 gcd(a, n). Show that the equation ax mod n 5 1 has a solution if and only if d 5 1. (This exercise is referred to in Chapter 2.) 14. Show that 5n 1 3 and 7n 1 4 are relatively prime for all n. 15. Prove that every prime greater than 3 can be written in the form 6n 1 1 or 6n 1 5. 16. Determine 71000 mod 6 and 61001 mod 7. 17. Let a, b, s, and t be integers. If a mod st 5 b mod st, show that a mod s 5 b mod s and a mod t 5 b mod t. What condition on s and t is needed to make the converse true? (This exercise is referred to in Chapter 8.) 18. Determine 8402 mod 5. 19. Show that gcd(a, bc) 5 1 if and only if gcd(a, b) 5 1 and gcd(a, c) 5 1. (This exercise is referred to in Chapter 8.) 20. Let p1, p2, . . . , pn be primes. Show that p1 p2 ? ? ? pn 1 1 is divisible by none of these primes. 21. Prove that there are infinitely many primes. (Hint: Use Exercise 20.) 22. For every positive integer n, prove that 1 1 2 1 ? ? ? 1 n 5 n(n 1 1)/2. 23. For every positive integer n, prove that a set with exactly n elements has exactly 2n subsets (counting the empty set and the entire set). 24. For any positive integer n, prove that 2n 32n 2 1 is always divisible by 17. 25. Prove that there is some positive integer n such that n, n 1 1, n 1 2, ? ? ? , n 1 200 are all composite.

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26. (Generalized Euclid’s Lemma) If p is a prime and p divides a1a2 ? ? ? an, prove that p divides ai for some i. 27. Use the Generalized Euclid’s Lemma (see Exercise 26) to establish the uniqueness portion of the Fundamental Theorem of Arithmetic. 28. What is the largest bet that cannot be made with chips worth $7.00 and $9.00? Verify that your answer is correct with both forms of induction. 29. Prove that the First Principle of Mathematical Induction is a consequence of the Well Ordering Principle. 30. The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . . In general, the Fibonacci numbers are defined by f1 5 1, f2 5 1, and for n $ 3, fn 5 fn21 1 fn22. Prove that the nth Fibonacci number fn satisfies fn , 2n. 31. In the cut “As” from Songs in the Key of Life, Stevie Wonder mentions the equation 8 3 8 3 8 3 8 5 4. Find all integers n for which this statement is true, modulo n. 32. Prove that for every integer n, n3 mod 6 5 n mod 6. 33. If it were 2:00 A.M. now, what time would it be 3736 hours from now? 34. Determine the check digit for a money order with identification number 7234541780. 35. Suppose that in one of the noncheck positions of a money order number, the digit 0 is substituted for the digit 9 or vice versa. Prove that this error will not be detected by the check digit. Prove that all other errors involving a single position are detected. 36. Suppose that a money order identification number and check digit of 21720421168 is erroneously copied as 27750421168. Will the check digit detect the error? 37. A transposition error involving distinct adjacent digits is one of the form . . . ab . . . → . . . ba . . . with a  b. Prove that the money order check digit scheme will not detect such errors unless the check digit itself is transposed. 38. Determine the check digit for the Avis rental car with identification number 540047. (See Example 6.) 39. Show that a substitution of a digit ai9 for the digit ai (ai9  ai) in a noncheck position of a UPS number is detected if and only if |a i 2 a i 9|  7. 40. Determine which transposition errors involving adjacent digits are detected by the UPS check digit. 41. Use the UPC scheme to determine the check digit for the number 07312400508.

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42. Explain why the check digit for a money order for the number N is the repeated decimal digit in the real number N 4 9. 43. The 10-digit International Standard Book Number (ISBN-10) a1a2a3a4a5a6a7a8 a9a10 has the property (a1, a2, . . . , a10) ? (10, 9, 8, 7, 6, 5, 4, 3, 2, 1) mod 11 5 0. The digit a10 is the check digit. When a10 is required to be 10 to make the dot product 0, the character X is used as the check digit. Verify the check digit for the ISBN-10 assigned to this book. 44. Suppose that an ISBN-10 has a smudged entry where the question mark appears in the number 0-716?-2841-9. Determine the missing digit. 45. Suppose three consecutive digits abc of an ISBN-10 are scrambled as bca. Which such errors will go undetected? 46. The ISBN-10 0-669-03925-4 is the result of a transposition of two adjacent digits not involving the first or last digit. Determine the correct ISBN-10. 47. Suppose the weighting vector for ISBN-10s was changed to (1, 2, 3, 4, 5, 6, 7, 8, 9, 10). Explain how this would affect the check digit. 48. Use the two-check-digit error-correction method described in this chapter to append two check digits to the number 73445860. 49. Suppose that an eight-digit number has two check digits appended using the error-correction method described in this chapter and it is incorrectly transcribed as 4302511568. If exactly one digit is incorrect, determine the correct number. 50. The state of Utah appends a ninth digit a9 to an eight-digit driver’s license number a1a2 . . . a8 so that (9a1 1 8a2 1 7a3 1 6a4 1 5a5 1 4a6 1 3a7 1 2a8 1 a9) mod 10 5 0. If you know that the license number 149105267 has exactly one digit incorrect, explain why the error cannot be in position 2, 4, 6, or 8. 51. Complete the proof of Theorem 0.7. 52. Let S be the set of real numbers. If a, b [ S, define a , b if a 2 b is an integer. Show that , is an equivalence relation on S. Describe the equivalence classes of S. 53. Let S be the set of integers. If a, b [ S, define aRb if ab $ 0. Is R an equivalence relation on S? 54. Let S be the set of integers. If a, b [ S, define aRb if a 1 b is even. Prove that R is an equivalence relation and determine the equivalence classes of S. 55. Complete the proof of Theorem 0.6 by showing that , is an equivalence relation on S.

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56. Prove that none of the integers 11, 111, 1111, 11111, . . . is a square of an integer. 57. (Cancellation Property) Suppose a, b and g are functions. If ag 5 bg and g is one-to-one and onto, prove that a 5 b.

Computer Exercises There is nothing more practical than a good theory. LEONID BREZHNEV

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software checks the validity of a Postal Service money order number. Use it to verify that 39539881642 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with the 9 in position 2 replaced with a 0. Was the error detected? Explain why or why not. Enter the number with two digits transposed. Was the error detected? Explain why or why not. 2. This software checks the validity of a UPC number. Use it to verify that 090146003386 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with two consecutive digits transposed. Was the error detected? Enter the number with the second 3 and the 8 transposed. Was the error detected? Explain why or why not. Enter the number with the 9 and the 1 transposed. Was the error detected? Explain why or why not. 3. This software checks the validity of a UPS number. Use it to verify that 8733456723 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with two consecutive digits transposed. Was the error detected? Enter the number with the 8 replaced by 1. Was the error detected? Explain why or why not. 4. This software checks the validity of an identification number on a bank check. Use it to verify that 091902049 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with two consecutive digits transposed. Was the error detected? Enter the number with the 2 and the 4 transposed. Was the error detected? Explain why or why not. 5. This software checks the validity of an ISBN-10. Use it to verify that 0395872456 is valid. Now enter the same number with one digit incorrect. Was the error detected? Enter the number with two digits transposed (they need not be consecutive). Was the error detected?

26

Integers and Equivalence Relations

6. This software determines the two check digits for the mod 11 decimal error-correction scheme discussed in this chapter. Run the program with the input 21355432, 20965744, 10033456. Then enter these numbers with the two check digits appended with one digit incorrect. Was the error corrected?

Suggested Readings Linda Deneen, “Secret Encryption with Public Keys,” The UMAP Journal 8 (1987): 9–29. This well-written article describes several ways in which modular arithmetic can be used to code secret messages. They range from a simple scheme used by Julius Caesar to a highly sophisticated scheme invented in 1978 and based on modular n arithmetic, where n has more than 200 digits. J. A. Gallian, “Assigning Driver’s License Numbers,” Mathematics Magazine 64 (1991): 13–22. This article describes various methods used by the states to assign driver’s license numbers. Several include check digits for error detection. This article can be downloaded at http://www.d.umn.edu/~jgallian/ license.pdf J. A. Gallian, “The Mathematics of Identification Numbers,” The College Mathematics Journal 22 (1991): 194–202. This article is a comprehensive survey of check digit schemes that are associated with identification numbers. This article can be downloaded at http://www.d.umn.edu/~jgallian/ident.pdf J. A. Gallian and S. Winters, “Modular Arithmetic in the Marketplace,” The American Mathematical Monthly 95 (1988): 548–551. This article provides a more detailed analysis of the check digit schemes presented in this chapter. In particular, the error detection rates for the various schemes are given. This article can be downloaded at http://www.d.umn.edu/~jgallian/marketplace.pdf

PA R T

2

Groups

For online student resources, visit this textbook’s website at http://college.hmco.com/PIC/gallian7e

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Introduction 1 to Groups Symmetry is a vast subject, significant in art and nature. Mathematics lies at its root, and it would be hard to find a better one on which to demonstrate the working of the mathematical intellect. HERMANN WEYL, Symmetry

Symmetries of a Square Suppose we remove a square region from a plane, move it in some way, then put the square back into the space it originally occupied. Our goal in this chapter is to describe in some reasonable fashion all possible ways in which this can be done. More specifically, we want to describe the possible relationships between the starting position of the square and its final position in terms of motions. However, we are interested in the net effect of a motion, rather than in the motion itself. Thus, for example, we consider a 908 rotation and a 4508 rotation as equal, since they have the same net effect on every point. With this simplifying convention, it is an easy matter to achieve our goal. To begin, we can think of the square region as being transparent (glass, say), with the corners marked on one side with the colors blue, white, pink, and green. This makes it easy to distinguish between motions that have different effects. With this marking scheme, we are now in a position to describe, in simple fashion, all possible ways in which a square object can be repositioned. See Figure 1.1. We now claim that any motion—no matter how complicated—is equivalent to one of these eight. To verify this claim, observe that the final position of the square is completely determined by the location and orientation (that is, face up or face down) of any particular corner. But, clearly, there are only four locations and two orientations for a given corner, so there are exactly eight distinct final positions for the corner.

29

30

Groups

P

W

G

B

P

W

R90 = Rotation of 90° (counterclockwise) G

B

P

W R180

G

P

R270 = Rotation of 270°

= Flip about a horizontal axis

D

B

W

G

B

P

W

G

B

W

B

P

G

B

G

W

P

G

P

B

W

G

B

P

W

W

P

B

G

P

G

W

B

B

W

G

P

R270

H G

V

W

R90

R180 = Rotation of 180°

H

P R0

R0 = Rotation of 0° (no change in position)

P

= Flip about a vertical axis

= Flip about the main diagonal

D = Flip about the other diagonal

B W

G

B

P

W

G

B

P

W

G

V

D

D

B

Figure 1.1

Let’s investigate some consequences of the fact that every motion is equal to one of the eight listed in Figure 1.1. Suppose a square is repositioned by a rotation of 908 followed by a flip about the horizontal axis of symmetry. In pictures, P G

W B

R90

W

B H

P

G

P

G

W

B

Thus, we see that this pair of motions—taken together—is equal to the single motion D. This observation suggests that we can compose two motions to obtain a single motion. And indeed we can, since the

31

1 | Introduction to Groups

eight motions may be viewed as functions from the square region to itself, and as such we can combine them using function composition. With this in mind, we may now write HR90 5 D. The eight motions R0, R90, R180, R270, H, V, D, and D9, together with the operation composition, form a mathematical system called the dihedral group of order 8 (the order of a group is the number of elements it contains). It is denoted by D4. Rather than introduce the formal definition of a group here, let’s look at some properties of groups by way of the example D4. To facilitate future computations, we construct an operation table or Cayley table (so named in honor of the prolific English mathematician Arthur Cayley, who first introduced them in 1854) for D4 below. The circled entry represents the fact that D 5 HR90. (In general, ab denotes the entry at the intersection of the row with a at the left and the column with b at the top.)

R0 R90 R180 R270 H V D D9

R0

R90

R180

R270

H

V

D

R0 R90 R180 R270 H V D D9

R90 R180 R270 R0 D  D9 V H

R180 R270 R0 R90 V H D9 D

R270 R0 R90 R180 D9 D H V

H D9 V D R0 R180 R270 R90

V D H D9 R180 R0 R90 R270

D H D9 V R90 R270 R0 R180

D9 D9 V D H R270 R90 R180 R0

Notice how orderly this table looks! This is no accident. Perhaps the most important feature of this table is that it has been completely filled in without introducing any new motions. Of course, this is because, as we have already pointed out, any sequence of motions turns out to be the same as one of these eight. Algebraically, this says that if A and B are in D4, then so is AB. This property is called closure, and it is one of the requirements for a mathematical system to be a group. Next, notice that if A is any element of D4, then AR0 5 R0 A 5 A. Thus, combining any element A on either side with R0 yields A back again. An element R0 with this property is called an identity, and every group must have one. Moreover, we see that for each element A in D4, there is exactly one element B in D4 such that AB 5 BA 5 R0. In this case, B is said to be the inverse of A and vice versa. For example, R90 and R270 are inverses of each other, and H is its own inverse. The term inverse is a descriptive one, for if A and B are inverses of each other, then B “undoes” whatever A “does,” in the sense that A and B taken together in either order produce R0, representing no change. Another striking feature

32

Groups

of the table is that every element of D4 appears exactly once in each row and column. This feature is something that all groups must have, and, indeed, it is quite useful to keep this fact in mind when constructing the table in the first place. Another property of D4 deserves special comment. Observe that HD Z DH but R90R180 5 R180R90. Thus, in a group, ab may or may not be the same as ba. If it happens that ab 5 ba for all choices of group elements a and b, we say the group is commutative or—better yet— Abelian (in honor of the great Norwegian mathematician Niels Abel). Otherwise, we say the group is non-Abelian. Thus far, we have illustrated, by way of D4, three of the four conditions that define a group—namely, closure, existence of an identity, and existence of inverses. The remaining condition required for a group is associativity; that is, (ab)c 5 a(bc) for all a, b, c in the set. To be sure that D4 is indeed a group, we should check this equation for each of the 83 5 512 possible choices of a, b, and c in D4. In practice, however, this is rarely done! Here, for example, we simply observe that the eight motions are functions and the operation is function composition. Then, since function composition is associative, we do not have to check the equations.

The Dihedral Groups The analysis carried out above for a square can similarly be done for an equilateral triangle or regular pentagon or, indeed, any regular n-gon (n $ 3). The corresponding group is denoted by Dn and is called the dihedral group of order 2n. The dihedral groups arise frequently in art and nature. Many of the decorative designs used on floor coverings, pottery, and buildings have one of the dihedral groups as a group of symmetry. Corporation logos are rich sources of dihedral symmetry [1]. Chrysler’s logo has D5 as a symmetry group, and that of Mercedes-Benz has D3. The ubiquitous five-pointed star has symmetry group D5. The phylum Echinodermata contains many sea animals (such as starfish, sea cucumbers, feather stars, and sand dollars) that exhibit patterns with D5 symmetry. Chemists classify molecules according to their symmetry. Moreover, symmetry considerations are applied in orbital calculations, in determining energy levels of atoms and molecules, and in the study of molecular vibrations. The symmetry group of a pyramidal molecule such as ammonia (NH3), depicted in Figure 1.2, has symmetry group D3.

1 | Introduction to Groups

33

N

H H H

Figure 1.2 A pyramidal molecule with symmetry group D3.

Mineralogists determine the internal structures of crystals (that is, rigid bodies in which the particles are arranged in three-dimensional repeating patterns—table salt and table sugar are two examples) by studying two-dimensional x-ray projections of the atomic makeup of the crystals. The symmetry present in the projections reveals the internal symmetry of the crystals themselves. Commonly occurring symmetry patterns are D4 and D6 (see Figure 1.3). Interestingly, it is mathematically impossible for a crystal to possess a Dn symmetry pattern with n 5 5 or n . 6.

Figure 1.3 X-ray diffraction photos revealing D4 symmetry patterns in crystals.

The dihedral group of order 2n is often called the group of symmetries of a regular n-gon. A plane symmetry of a figure F in a plane is a function from the plane to itself that carries F onto F and preserves distances; that is, for any points p and q in the plane, the distance from the image of p to the image of q is the same as the

34

Groups

distance from p to q. (The term symmetry is from the Greek word symmetros, meaning “of like measure.”) The symmetry group of a plane figure is the set of all symmetries of the figure. Symmetries in three dimensions are defined analogously. Obviously, a rotation of a plane about a point in the plane is a symmetry of the plane, and a rotation about a line in three dimensions is a symmetry in three-dimensional space. Similarly, any translation of a plane or of three-dimensional space is a symmetry. A reflection across a line L is that function that leaves every point of L fixed and takes any point q, not on L, to the point q9 so that L is the perpendicular bisector of the line segment joining q and q9 (see Figure 1.4). A reflection across a plane in three dimensions is defined analogously. Notice that the restriction of a 1808 rotation about a line L in three dimensions to a plane containing L is a reflection across L in the plane. Thus, in the dihedral groups, the motions that we described as flips about axes of symmetry in three dimensions (for example, H, V, D, D9) are reflections across lines in two dimensions. Just as a reflection across a line is a plane symmetry that cannot be achieved by a physical motion of the plane in two dimensions, a reflection across a plane is a three-dimensional symmetry that cannot be achieved by a physical motion of three-dimensional space. A cup, for instance, has reflective symmetry across the plane bisecting the cup, but this symmetry cannot be duplicated with a physical motion in three dimensions. q q9 L

Figure 1.4

Many objects and figures have rotational symmetry but not reflective symmetry. A symmetry group consisting of the rotational symmetries of 08, 3608/n, 2(3608)/n, . . . , (n 2 1)3608/n, and no other symmetries is called a cyclic rotation group of order n and is denoted by 7R360/n8. Cyclic rotation groups, along with dihedral groups, are favorites of artists, designers, and nature. Figure 1.5 illustrates with corporate logos the cyclic rotation groups of orders 2, 3, 4, 5, 6, 8, 16, and 20. Further examples of the occurrence of dihedral groups and cyclic groups in art and nature can be found in the references. A study of symmetry in greater depth is given in Chapters 27 and 28.

1 | Introduction to Groups

35

Figure 1.5 Logos with cyclic rotation symmetry groups.

Exercises The only way to learn mathematics is to do mathematics. PAUL HALMOS, Hilbert Space Problem Book

1. With pictures and words, describe each symmetry in D3 (the set of symmetries of an equilateral triangle). 2. Write out a complete Cayley table for D3. 3. Is D3 Abelian? 4. Describe in pictures or words the elements of D5 (symmetries of a regular pentagon). 5. For n $ 3, describe the elements of Dn. (Hint: You will need to consider two cases—n even and n odd.) How many elements does Dn have? 6. In Dn, explain geometrically why a reflection followed by a reflection must be a rotation. 7. In Dn, explain geometrically why a rotation followed by a rotation must be a rotation. 8. In Dn, explain geometrically why a rotation and a reflection taken together in either order must be a reflection. 9. Associate the number 11 with a rotation and the number 21 with a reflection. Describe an analogy between multiplying these two numbers and multiplying elements of Dn.

36

Groups

10. If r1, r2, and r3 represent rotations from Dn and f1, f2, and f3 represent reflections from Dn, determine whether r1r2 f1r3 f2 f3r3 is a rotation or a reflection. 11. Find elements A, B, and C in D4 such that AB 5 BC but A Z C. (Thus, “cross cancellation” is not valid.) 12. Explain what the following diagram proves about the group Dn. 1

1

n

F 2

n

R360 / n 2 2

1

2

n–1 2

R360/ n n

1

n

F 1

3

3

1

13. Describe the symmetries of a nonsquare rectangle. Construct the corresponding Cayley table. 14. Describe the symmetries of a parallelogram that is neither a rectangle nor a rhombus. Describe the symmetries of a rhombus that is not a rectangle. 15. Describe the symmetries of a noncircular ellipse. Do the same for a hyperbola. 16. Consider an infinitely long strip of equally spaced H’s: ???HHHH??? Describe the symmetries of this strip. Is the group of symmetries of the strip Abelian? 17. For each of the snowflakes in the figure, find the symmetry group and locate the axes of reflective symmetry (disregard imperfections).

Photographs of snowflakes from the Bentley and Humphrey atlas.

1 | Introduction to Groups

37

18. Determine the symmetry group of the outer shell of the cross section of the human immunodeficiency virus (HIV) shown below.

19. Does an airplane propeller have a cyclic symmetry group or a dihedral symmetry group? 20. Bottle caps that are pried off typically have 22 ridges around the rim. Find the symmetry group of such a cap. 21. What group theoretic property do upper-case letters F, G, J, K, L, P, Q, R have that is not shared by the remaining upper-case letters in the alphabet? 22. For each design below, determine the symmetry group (ignore imperfections).

38

Groups

23. What would the effect be if a six-bladed ceiling fan were designed so that the centerlines of two of the blades were at a 708 angle and all the other blades were set at a 588 angle?

Reference 1. B. B. Capitman, American Trademark Designs, New York: Dover, 1976.

Suggested Reading Michael Field and Martin Golubitsky, Symmetry in Chaos, Oxford University Press, 1992. This book has many beautiful symmetric designs that arise in chaotic dynamic systems.

Niels Abel He [Abel] has left mathematicians something to keep them busy for five hundred years. CHARLES HERMITE

A 500-kroner bank note first issued by Norway in 1948.

NIELS HENRIK ABEL, one of the foremost mathematicians of the 19th century, was born in Norway on August 5, 1802. At the age of 16, he began reading the classic mathematical works of Newton, Euler, Lagrange, and Gauss. When Abel was 18 years old, his father died, and the burden of supporting the family fell upon him. He took in private pupils and did odd jobs, while continuing to do mathematical research. At the age of 19, Abel solved a problem that had vexed leading mathematicians for hundreds of years. He proved that, unlike the situation for equations of degree 4 or less, there is no finite (closed) formula for the solution of the general fifth-degree equation. Although Abel died long before the advent of the subjects that now make up abstract algebra, his solution to the quintic problem laid the groundwork for many of these subjects. Just when his work was beginning to receive the attention it deserved, Abel contracted tuberculosis. He died on April 6, 1829, at the age of 26.

This stamp was issued in 1929 to commemorate the 100th anniversary of Abel’s death.

In recognition of the fact that there is no Nobel Prize for mathematics, in 2002 Norway established the Abel Prize as the “Nobel Prize in mathematics” in honor of its native son. At approximately the $1,000,000 level, the Abel Prize is now seen as an award equivalent to the Nobel Prize. To find more information about Abel, visit: http://www-groups.dcs.st-and .ac.uk/~history/

39

2 Groups

A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one. PAUL R. HALMOS

Definition and Examples of Groups The term group was used by Galois around 1830 to describe sets of one-to-one functions on finite sets that could be grouped together to form a set closed under composition. As is the case with most fundamental concepts in mathematics, the modern definition of a group that follows is the result of a long evolutionary process. Although this definition was given by both Heinrich Weber and Walter von Dyck in 1882, it did not gain universal acceptance until the 20th century. Definition Binary Operation

Let G be a set. A binary operation on G is a function that assigns each ordered pair of elements of G an element of G.

A binary operation on a set G, then, is simply a method (or formula) by which the members of an ordered pair from G combine to yield a new member of G. This condition is called closure. The most familiar binary operations are ordinary addition, subtraction, and multiplication of integers. Division of integers is not a binary operation on the integers because an integer divided by an integer need not be an integer. The binary operations addition modulo n and multiplication modulo n on the set {0, 1, 2, . . . , n 2 1}, which we denote by Zn, play an extremely important role in abstract algebra. In certain situations we will want to combine the elements of Zn by addition modulo n only; in other situations we will want to use both addition modulo n and multiplication modulo n to combine the elements. It will be clear

40

2 | Groups

41

from the context whether we are using addition only or addition and multiplication. For example, when multiplying matrices with entries from Zn, we will need both addition modulo n and multiplication modulo n. Definition Group

Let G be a set together with a binary operation (usually called multiplication) that assigns to each ordered pair (a, b) of elements of G an element in G denoted by ab. We say G is a group under this operation if the following three properties are satisfied. 1. Associativity. The operation is associative; that is, (ab)c 5 a(bc) for all a, b, c in G. 2. Identity. There is an element e (called the identity) in G such that ae 5 ea 5 a for all a in G. 3. Inverses. For each element a in G, there is an element b in G (called an inverse of a) such that ab 5 ba 5 e.

In words, then, a group is a set together with an associative operation such that there is an identity, every element has an inverse, and any pair of elements can be combined without going outside the set. Be sure to verify closure when testing for a group (see Example 5). Notice that if a is the inverse of b, then b is the inverse of a. If a group has the property that ab 5 ba for every pair of elements a and b, we say the group is Abelian. A group is non-Abelian if there is some pair of elements a and b for which ab 2 ba. When encountering a particular group for the first time, one should determine whether or not it is Abelian. Now that we have the formal definition of a group, our first job is to build a good stock of examples. These examples will be used throughout the text to illustrate the theorems. (The best way to grasp the meat of a theorem is to see what it says in specific cases.) As we progress, the reader is bound to have hunches and conjectures that can be tested against the stock of examples. To develop a better understanding of the following examples, the reader should supply the missing details. EXAMPLE 1 The set of integers Z (so denoted because the German word for numbers is Zahlen), the set of rational numbers Q (for quotient), and the set of real numbers R are all groups under ordinary addition. In each case, the identity is 0 and the inverse of a is 2a.

42

Groups

EXAMPLE 2 The set of integers under ordinary multiplication is not a group. Since the number 1 is the identity, property 3 fails. For example, there is no integer b such that 5b 5 1. EXAMPLE 3 The subset {1, 21, i, 2i} of the complex numbers is a group under complex multiplication. Note that 21 is its own inverse, whereas the inverse of i is 2i, and vice versa. EXAMPLE 4 The set Q1 of positive rationals is a group under ordinary multiplication. The inverse of any a is 1/a 5 a21. EXAMPLE 5 The set S of positive irrational numbers together with 1 under multiplication satisfies the three properties given in the definition of a group but is not a group. Indeed, "2 ? "2 5 2, so S is not closed under multiplication. a b EXAMPLE 6 A rectangular array of the form c d is called a c d 2 3 2 matrix. The set of all 2 3 2 matrices with real entries is a group under componentwise addition. That is, c

a1 b1 a2 b2 a1 1 a2 b1 1 b2 d 1 c d 5 c d c1 d1 c2 d2 c1 1 c2 d1 1 d2

The identity is c

0 0 a b 2a 2b d , and the inverse of c d is c d. 0 0 c d 2c 2d

EXAMPLE 7 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a group under addition modulo n. For any j . 0 in Zn, the inverse of j is n 2 j. This group is usually referred to as the group of integers modulo n. As we have seen, the real numbers, the 2 3 2 matrices with real entries, and the integers modulo n are all groups under the appropriate addition. But what about multiplication? In each case, the existence of some elements that do not have inverses prevents the set from being a group under the usual multiplication. However, we can form a group in each case by simply throwing out the rascals. Examples 8, 9, and 11 illustrate this. EXAMPLE 8 The set R* of nonzero real numbers is a group under ordinary multiplication. The identity is 1. The inverse of a is 1/a.

2 | Groups

43

a b d is the c d number ad 2 bc. If A is a 2 3 2 matrix, det A denotes the determinant of A. The set EXAMPLE 9† The determinant of the 2 3 2 matrix c

GL(2, R) 5 e c

a b d ` a, b, c, d c d

P R, ad

2 bc ? 0 f

of 2 3 2 matrices with real entries and nonzero determinant is a nonAbelian group under the operation c

a a 1 b1c2 a1b2 1 b1d2 a1 b1 a2 b2 dc d 5 c 1 2 d c1 d1 c2 d2 c1a2 1 d1c2 c1b2 1 d1d2 .

The first step in verifying that this set is a group is to show that the product of two matrices with nonzero determinant also has nonzero determinant. This follows from the fact that for any pair of 2 3 2 matrices A and B, det (AB) 5 (det A)(det B). Associativity can be verified by direct (but cumbersome) calcula1 0 a b tions. The identity is c d ; the inverse of c d is 0 1 c d d 2b ad 2 bc ad 2 bc ≥ ¥ 2c a ad 2 bc ad 2 bc (explaining the requirement that ad 2 bc 2 0). This very important non-Abelian group is called the general linear group of 2 3 2 matrices over R. EXAMPLE 10 The set of all 2 3 2 matrices with real number entries is not a group under the operation defined in Example 9. Inverses do not exist when the determinant is 0. Now that we have shown how to make subsets of the real numbers and subsets of the set of 2 3 2 matrices into multiplicative groups, we next consider the integers under multiplication modulo n. †For

simplicity, we have restricted our matrix examples to the 2 3 2 case. However, readers who have had linear algebra can readily generalize to n 3 n matrices.

44

Groups

EXAMPLE 11 (L. Euler, 1761) By Exercise 13 in Chapter 0, an integer a has a multiplicative inverse modulo n if and only if a and n are relatively prime. So, for each n . 1, we define U(n) to be the set of all positive integers less than n and relatively prime to n. Then U(n) is a group under multiplication modulo n. (We leave it to the reader to check that this set is closed under this operation.) For n 5 10, we have U(10) 5 {1, 3, 7, 9}. The Cayley table for U(10) is mod 10

1

3

7

9

1 3 7 9

1 3 7 9

3 9 1 7

7 1 9 3

9 7 3 1

(Recall that ab mod n is the unique integer r with the property a ? b 5 nq 1 r, where 0 # r , n and a ? b is ordinary multiplication.) In the case that n is a prime, U(n) 5 {1, 2, . . . , n 2 1}. In his classic book Lehrbuch der Algebra, published in 1899, Heinrich Weber gave an extensive treatment of the groups U(n) and described them as the most important examples of finite Abelian groups. EXAMPLE 12 The set {0, 1, 2, 3} is not a group under multiplication modulo 4. Although 1 and 3 have inverses, the elements 0 and 2 do not. EXAMPLE 13 The set of integers under subtraction is not a group, since the operation is not associative. With the examples given thus far as a guide, it is wise for the reader to pause here and think of his or her own examples. Study actively! Don’t just read along and be spoon-fed by the book. EXAMPLE 14 For all integers n $ 1, the set of complex nth roots of unity e cos

k ? 360° k ? 360° 1 i sin ` k 5 0, 1, 2, . . . , n 2 1 f n n

(i.e., complex zeros of xn 2 1) is a group under multiplication. (See DeMoivre’s Theorem—Example 7 in Chapter 0.) Compare this group with the one in Example 3.

2 | Groups

45

The complex number a 1 bi can be represented geometrically as the point (a, b) in a plane coordinatized by a horizontal real axis and a vertical i or imaginary axis. The distance from the point a 1 bi to the origin is "a 2 1b 2 and is often denoted by Ua 1 bi|. For any angle u, the line segment joining the complex number cos u 1 i sin u and the origin forms an angle of u with the positive real axis. Thus, the six complex zeros of x6 5 1 are located at points around the circle of radius 1, 60° apart, as shown in Figure 2.1. Imaginary – 1 + √3 i 2 2

1 + √3 2 2 i

√3 2 60°

–1

1 1 2

– 1 – √3 i 2 2

Real

1 – √3 2 2 i

Figure 2.1

EXAMPLE 15 The set Rn 5 {(a1, a2, . . . , an) U a1, a2, . . . , an [ R} is a group under componentwise addition [i.e., (a1, a2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn)]. EXAMPLE 16 For a fixed point (a, b) in R2, define Ta,b: R2 → R2 by (x, y) → (x 1 a, y 1 b). Then G 5 {Ta,b U a, b [ R} is a group under function composition. Straightforward calculations show that Ta,bTc,d 5 Ta1c,b1d. From this formula we may observe that G is closed, T0,0 is the identity, the inverse of Ta,b is T2a,2b, and G is Abelian. Function composition is always associative. The elements of G are called translations. EXAMPLE 17 The set of all 2 3 2 matrices with determinant 1 with entries from Q (rationals), R (reals), C (complex numbers), or Zp (p a prime) is a non-Abelian group under matrix multiplication. This group is called the special linear group of 2 3 2 matrices over Q, R, C, or Zp, respectively.

46

Groups

If the entries are from F, where F is any of the above, we denote this group by SL(2, F). For the group SL(2, F), the formula given in Example 9 for a b d 2b the inverse of c d simplifies to c d . When the matrix c d 2c a entries are from Zp, we use modulo p arithmetic to compute determinants, matrix products, and inverses. To illustrate the case SL(2, Z5), 3 4 consider the element A 5 c d . Then det A 5 (3 ? 4 2 4 ? 4) mod 5 5 4 4 4 24 4 1 24 mod 5 5 1, and the inverse of A is c d 5 c d. Note 24 3 1 3 3 4 4 1 1 0 that c dc d 5 c d when the arithmetic is done modulo 5. 4 4 1 3 0 1 Example 9 is a special case of the following general construction. EXAMPLE 18 Let F be any of Q, R, C, or Zp ( p a prime). The set GL(2, F) of all 2 3 2 matrices with nonzero determinants and entries from F is a non-Abelian group under matrix multiplication. As in Example 17, when F is Zp, modulo p arithmetic is used to calculate determinants, the matrix products, and inverses. The formula given in a b d remains valid for elements from Example 9 for the inverse of c c d GL(2, Zp) provided we interpret division by ad 2 bc as multiplication by the inverse of ad 2 bc modulo p. For example, in GL(2, Z7), 4 5 d . Then the determinant (ad 2 bc) mod 7 is (12 2 30) consider c 6 3 mod 7 5 218 mod 7 5 3 and the inverse of 3 is 5 [since (3 ? 5) 4 5 3?5 2?5 1 3 d is c d 5 c d. mod 7 5 1]. So, the inverse of c 6 3 1?5 4?5 5 6 4 5 1 3 1 0 d c d 5 c d in GL(2, Z7)]. [The reader should check that c 6 3 5 6 0 1 EXAMPLE 19 The set {1, 2, . . . , n 2 1} is a group under multiplication modulo n if and only if n is prime. EXAMPLE 20 The set of all symmetries of the infinite ornamental pattern in which arrowheads are spaced uniformly a unit apart along

47

2 | Groups

a line is an Abelian group under composition. Let T denote a translation to the right by one unit, T 21 a translation to the left by one unit, and H a reflection across the horizontal line of the figure. Then, every member of the group is of the form x1x2 ? ? ? xn, where each xi [ {T, T 21, H}. In this case, we say that T, T 21, and H generate the group. Table 2.1 summarizes many of the specific groups that we have presented thus far. As the examples above demonstrate, the notion of a group is a very broad one indeed. The goal of the axiomatic approach is to find properties general enough to permit many diverse examples having these properties and specific enough to allow one to deduce many interesting consequences. The goal of abstract algebra is to discover truths about algebraic systems (that is, sets with one or more binary operations) that are independent of the specific nature of the operations. All one knows or needs to know is that these operations, whatever they may be, have Table 2.1 Summary of Group Examples (F can be any of Q, R, C, or Zp; L is a reflection) Group

Operation

Z Q1

Addition Multiplication

0 1

Zn R* GL(2, F)

Addition mod n Multiplication Matrix multiplication

0 1

U(n) Rn SL(2, F)

Dn

Multiplication mod n Componentwise addition Matrix multiplication Composition

Identity

c

1 0

0 d 1

Form of Element k m/n, m, n . 0 k x a b c d c ,d

ad 2 bc 2 0 k, gcd(k, n) 5 1 (0, 0, …, 0) (a1, a2, …, an) 1

c

1 0

0 d 1 R0

a b d c d ad 2 bc 5 1 Ra, L c

Inverse

Abelian

2k n/m

Yes Yes

n2k 1/x

Yes Yes

d 2b ad 2 bc ad 2 bc ≥ ¥ 2c a ad 2 bc ad 2 bc Solution to kx mod n 5 1 (2a1, 2a2, …, 2an)

No Yes Yes

2b d a

No

R360 2 a, L

No

d c 2c

48

Groups

certain properties. We then seek to deduce consequences of these properties. This is why this branch of mathematics is called abstract algebra. It must be remembered, however, that when a specific group is being discussed, a specific operation must be given (at least implicitly).

Elementary Properties of Groups Now that we have seen many diverse examples of groups, we wish to deduce some properties that they share. The definition itself raises some fundamental questions. Every group has an identity. Could a group have more than one? Every group element has an inverse. Could an element have more than one? The examples suggest not. But examples can only suggest. One cannot prove that every group has a unique identity by looking at examples, because each example inherently has properties that may not be shared by all groups. We are forced to restrict ourselves to the properties that all groups have; that is, we must view groups as abstract entities rather than argue by example. The next three theorems illustrate the abstract approach. Theorem 2.1 Uniqueness of the Identity In a group G, there is only one identity element.

PROOF Suppose both e and e9 are identities of G. Then, 1. ae 5 a for all a in G, and 2. e9a 5 a for all a in G. The choices of a 5 e9 in (1) and a 5 e in (2) yield e9e 5 e9 and e9e 5 e. Thus, e and e9 are both equal to e9e and so are equal to each other. Because of this theorem, we may unambiguously speak of “the identity” of a group and denote it by “e” (because the German word for identity is Einheit). Theorem 2.2 Cancellation In a group G, the right and left cancellation laws hold; that is, ba 5 ca implies b 5 c, and ab 5 ac implies b 5 c.

2 | Groups

49

PROOF Suppose ba 5 ca. Let a9 be an inverse of a. Then, multiplying on the right by a9 yields (ba)a9 5 (ca)a9. Associativity yields b(aa9) 5 c(aa9). Then, be 5 ce and, therefore, b 5 c as desired. Similarly, one can prove that ab 5 ac implies b 5 c by multiplying by a9 on the left. A consequence of the cancellation property is the fact that in a Cayley table for a group, each group element occurs exactly once in each row and column (see Exercise 23). Another consequence of the cancellation property is the uniqueness of inverses. Theorem 2.3 Uniqueness of Inverses For each element a in a group G, there is a unique element b in G such that ab 5 ba 5 e.

PROOF Suppose b and c are both inverses of a. Then ab 5 e and ac 5 e, so that ab 5 ac. Canceling the a on both sides gives b 5 c, as desired. As was the case with the identity element, it is reasonable, in view of Theorem 2.3, to speak of “the inverse” of an element g of a group; in fact, we may unambiguously denote it by g21. This notation is suggested by that used for ordinary real numbers under multiplication. Similarly, when n is a positive integer, the associative law allows us to use gn to denote the unambiguous product gg ? ? ? g. n factors

g0

We define 5 e. When n is negative, we define gn 5 (g21)|n| [for example, g23 5 (g21)3]. Unlike for real numbers, in an abstract group we do not permit noninteger exponents such as g1/2. With this notation, the familiar laws of exponents hold for groups; that is, for all integers m and n and any group element g, we have gmgn 5 gm1n and (gm)n 5 gmn. Although the way one manipulates the group expressions gmgn and (gm)n coincides with the laws of exponents for real numbers, the laws of exponents fail to hold for expressions involving two group elements. Thus, for groups in general, (ab)n Z anbn (see Exercise 15). Also, one must be careful with this notation when dealing with a specific group whose binary operation is addition and is denoted by

50

Groups

“1.” In this case, the definitions and group properties expressed in multiplicative notation must be translated to additive notation. For example, the inverse of g is written as 2g. Likewise, for example, g3 Table 2.2 Multiplicative Group

Additive Group

a ? b or ab e or 1 a21 an

Multiplication Identity or one Multiplicative inverse of a Power of a

a1b 0 2a na

Addition Zero Additive inverse of a Multiple of a

ab21

Quotient

a2b

Difference

means g 1 g 1 g and is usually written as 3g, whereas g23 means (2g) 1 (2g) 1 (2g) and is written as 23g. When additive notation is used, do not interpret “ng” as combining n and g under the group operation; n may not even be an element of the group! Table 2.2 shows the common notation and corresponding terminology for groups under multiplication and groups under addition. As is the case for real numbers, we use a 2 b as an abbreviation for a 1 (2b). Because of the associative property, we may unambiguously write the expression abc, for this can be reasonably interpreted as only (ab)c or a(bc), which are equal. In fact, by using induction and repeated application of the associative property, one can prove a general associative property that essentially means that parentheses can be inserted or deleted at will without affecting the value of a product involving any number of group elements. Thus, a2(bcdb2) 5 a2b(cd )b2 5 (a2b)(cd )b2 5 a(abcdb)b, and so on. Although groups do not have the property that (ab)n 5 a nb n there is a simple relationship between (ab)21 and a21 and b21. Theorem 2.4 Socks-Shoes Property For group elements a and b, (ab)21 5 b21a21.

PROOF Since (ab)(ab)21 5 e and (ab)(b21a21) 5 a(bb21)a21 5 aea21 5 aa21 5 e, we have by Theorem 2.3 that (ab)21 5 b21a21.

2 | Groups

51

Historical Note We conclude this chapter with a bit of history concerning the noncommutativity of matrix multiplication. In 1925, quantum theory was replete with annoying and puzzling ambiguities. It was Werner Heisenberg who recognized the cause. He observed that the product of the quantum-theoretical analogs of the classical Fourier series did not necessarily commute. For all his boldness, this shook Heisenberg. As he later recalled [2, p. 94]: In my paper the fact that XY was not equal to YX was very disagreeable to me. I felt this was the only point of difficulty in the whole scheme, otherwise I would be perfectly happy. But this difficulty had worried me and I was not able to solve it.

Heisenberg asked his teacher, Max Born, if his ideas were worth publishing. Born was fascinated and deeply impressed by Heisenberg’s new approach. Born wrote [1, p. 217]: After having sent off Heisenberg’s paper to the Zeitschrift für Physik for publication, I began to ponder over his symbolic multiplication, and was soon so involved in it that I thought about it for the whole day and could hardly sleep at night. For I felt there was something fundamental behind it, the consummation of our endeavors of many years. And one morning, about the 10 July 1925, I suddenly saw light: Heisenberg’s symbolic multiplication was nothing but the matrix calculus, wellknown to me since my student days from Rosanes’ lectures in Breslau.

Born and his student, Pascual Jordan, reformulated Heisenberg’s ideas in terms of matrices, but it was Heisenberg who was credited with the formulation. In his autobiography, Born laments [1, p. 219]: Nowadays the textbooks speak without exception of Heisenberg’s matrices, Heisenberg’s commutation law, and Dirac’s field quantization. In fact, Heisenberg knew at that time very little of matrices and had to study them.

Upon learning in 1933 that he was to receive the Nobel Prize with Dirac and Schrödinger for this work, Heisenberg wrote to Born [1, p. 220]: If I have not written to you for such a long time, and have not thanked you for your congratulations, it was partly because of my rather bad conscience with respect to you. The fact that I am to receive the Nobel Prize alone, for work done in Göttingen in collaboration—you, Jordan, and I—this fact depresses me and I hardly know what to write to you. I am, of course, glad that our common efforts are now appreciated, and I enjoy the recollection of the beautiful time of collaboration. I also believe that all good physicists know how great was your and Jordan’s contribution to the structure of quantum mechanics—and this remains unchanged by a wrong decision from outside. Yet I myself can do nothing but thank you again for all the fine collaboration, and feel a little ashamed.

The story has a happy ending, however, because Born received the Nobel Prize in 1954 for his fundamental work in quantum mechanics.

52

Groups

Exercises “For example,” is not proof. Jewish Proverb

1. Give two reasons why the set of odd integers under addition is not a group. 2. Referring to Example 13, verify the assertion that subtraction is not associative. 3. Show that {1, 2, 3} under multiplication modulo 4 is not a group but that {1, 2, 3, 4} under multiplication modulo 5 is a group. 4. Show that the group GL(2, R) of Example 9 is non-Abelian by exhibiting a pair of matrices A and B in GL(2, R) such that AB 2 BA. 2 6 5. Find the inverse of the element c d in GL(2, Z11). 3 5 6. Give an example of group elements a and b with the property that a21ba 2 b. 7. Translate each of the following multiplicative expressions into its additive counterpart. Assume that the operation is commutative. a. a2b3 b. a22(b21c)2 c. (ab2)23c2 5 e 8. Show that the set {5, 15, 25, 35} is a group under multiplication modulo 40. What is the identity element of this group? Can you see any relationship between this group and U(8)?

Text not available due to copyright restrictions

10. List the members of H 5 {x 2 | x [ D4} and K 5 {x [ D4 | x 2 5 e}. 11. Prove that the set of all 2 3 2 matrices with entries from R and determinant 11 is a group under matrix multiplication. 12. For any integer n . 2, show that there are at least two elements in U(n) that satisfy x2 5 1.

2 | Groups

53

13. An abstract algebra teacher intended to give a typist a list of nine integers that form a group under multiplication modulo 91. Instead, one of the nine integers was inadvertently left out, so that the list appeared as 1, 9, 16, 22, 53, 74, 79, 81. Which integer was left out? (This really happened!) 14. Let G be a group with the following property: Whenever a, b, and c belong to G and ab 5 ca, then b 5 c. Prove that G is Abelian. (“Cross cancellation” implies commutativity.) 15. (Law of Exponents for Abelian Groups) Let a and b be elements of an Abelian group and let n be any integer. Show that (ab)n 5 anbn. Is this also true for non-Abelian groups? 16. (Socks-Shoes Property) Draw an analogy between the statement (ab)21 5 b21a21 and the act of putting on and taking off your socks and shoes. Find an example that shows that in a group, it is possible to have (ab)22 2 b22a22. Find distinct nonidentity elements a and b from a non-Abelian group such that (ab)21 5 a21b21. 17. Prove that a group G is Abelian if and only if (ab)21 5 a21b21 for all a and b in G. 18. Prove that in a group, (a21)21 5 a for all a. 19. For any elements a and b from a group and any integer n, prove that (a21ba)n 5 a21bna. 20. If a1, a2, . . . , an belong to a group, what is the inverse of a1a2 . . . an? 21. The integers 5 and 15 are among a collection of 12 integers that form a group under multiplication modulo 56. List all 12. 22. Give an example of a group with 105 elements. Give two examples of groups with 44 elements. 23. Prove that every group table is a Latin square†; that is, each element of the group appears exactly once in each row and each column. (This exercise is referred to in this chapter.) 24. Construct a Cayley table for U(12). 25. Suppose the table below is a group table. Fill in the blank entries.

†Latin

e

a

b

c

d

e a b c

e — — —

— b c d

— — d —

— — e a

— e — b

d

—

—

—

—

—

squares are useful in designing statistical experiments. There is also a close connection between Latin squares and finite geometries.

54

Groups

26. Prove that if (ab)2 5 a2b2 in a group G, then ab 5 ba. 27. Let a, b, and c be elements of a group. Solve the equation axb 5 c for x. Solve a21xa 5 c for x. 28. Prove that the set of all rational numbers of the form 3m6n, where m and n are integers, is a group under multiplication. 29. Let G be a finite group. Show that the number of elements x of G such that x3 5 e is odd. Show that the number of elements x of G such that x2 2 e is even. 30. Give an example of a group with elements a, b, c, d, and x such that axb 5 cxd but ab 2 cd. (Hence “middle cancellation” is not valid in groups.) 31. Let R be any rotation in some dihedral group and F any reflection in the same group. Prove that RFR 5 F. 32. Let R be any rotation in some dihedral group and F, any reflection in the same group. Prove that FRF 5 R21 for all integers k. 33. Suppose that G is a group with the property that for every choice of elements in G, axb 5 cxd implies ab 5 cd. Prove that G is Abelian. (“Middle cancellation” implies commutativity.) 34. In the dihedral group Dn, let R 5 R360/n and let F be any reflection. Write each of the following products in the form Ri or RiF, where 0 # i , n. a. In D4, FR22FR5 b. In D5, R23FR4FR22 c. In D6, FR5FR22F 35. Prove that if G is a group with the property that the square of every element is the identity, then G is Abelian. (This exercise is referred to in Chapter 26.) 36. Prove that the set of all 3 3 3 matrices with real entries of the form 1 £0 0

a 1 0

b c§ 1

is a group. (Multiplication is defined by 1 £0 0

a 1 0

1 b 1 a9 b9 c § £ 0 1 c9 § 5 £0 0 1 0 0 1

a 1 a9 1 0

b9 1 ac9 1 b c9 1 c §. 1

This group, sometimes called the Heisenberg group after the Nobel Prize–winning physicist Werner Heisenberg, is intimately related to the Heisenberg Uncertainty Principle of quantum physics.)

2 | Groups

55

37. Prove the assertion made in Example 19 that the set {1, 2, . . . , n 2 1} is a group under multiplication modulo n if and only if n is prime. 38. In a finite group, show that the number of nonidentity elements that satisfy the equation x5 5 e is a multiple of 4. If the stipulation that the group be finite is omitted, what can you say about the number of nonidentity elements that satisfy the equation x 5 5 e? a a 39. Let G 5 e c d 0 a P R, a ? 0 f . Show that G is a group under a a matrix multiplication. Explain why each element of G has an inverse even though the matrices have 0 determinant. (Compare with Example 10.)

Computer Exercises Almost immediately after the war, Johnny [Von Neumann] and I also began to discuss the possibilities of using computers heuristically to try to obtain insights into questions of pure mathematics. By producing examples and by observing the properties of special mathematical objects, one could hope to obtain clues as to the behavior of general statements which have been tested on examples. S. M. ULAM, Adventures of a Mathematician

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software prints the elements of U(n) and the inverse of each element. 2. This software determines the size of U(k). Run the program for k 5 9, 27, 81, 243, 25, 125, 49, 121. On the basis of this output, try to guess a formula for the size of U( pn) as a function of the prime p and the integer n. Run the program for k 5 18, 54, 162, 486, 50, 250, 98, 242. Make a conjecture about the relationship between the size of U(2pn) and the size of U( pn), where p is a prime greater than 2. 3. This software computes the inverse of any element in GL(2, Zp), where p is a prime. 4. This software determines the number of elements in GL(2, Zp) and SL(2, Zp). (The technical term for the number of elements in a group is the order of the group.) Run the program for p 5 3, 5, 7, and 11.

56

Groups

Do you see a relationship between the orders of GL (2, Zp) and SL(2, Zp) and p 2 1? Does this relationship hold for p 5 2? Based on these examples, does it appear that p always divides the order of SL (2, Z p)? What about p 2 1? What about p 1 1? Guess a formula for the order of SL (2, Zp). Guess a formula for the order of GL(2, Zp).

References 1. Max Born, My Life: Recollections of a Nobel Laureate, New York: Charles Scribner’s Sons, 1978. 2. J. Mehra and H. Rechenberg, The Historical Development of Quantum Theory, Vol. 3, New York: Springer-Verlag, 1982.

Suggested Readings Marcia Ascher, Ethnomathematics, Pacific Grove, CA: Brooks/Cole, 1991. Chapter 3 of this book describes how the dihedral group of order 8 can be used to encode the social structure of the kin system of family relationships among a tribe of native people of Australia. Arie Bialostocki, “An Application of Elementary Group Theory to Central Solitaire,” The College Mathematics Journal, May 1998: 208–212. The author uses properties of groups to analyze the peg board game central solitaire (which also goes by the name peg solitaire). J. E. White, “Introduction to Group Theory for Chemists,” Journal of Chemical Education 44 (1967): 128–135. Students interested in the physical sciences may find this article worthwhile. It begins with easy examples of groups and builds up to applications of group theory concepts and terminology to chemistry.

Finite Groups; 3 Subgroups In our own time, in the period 1960–1980, we have seen particle physics emerge as the playground of group theory. FREEMAN DYSON

Terminology and Notation As we will soon discover, finite groups—that is, groups with finitely many elements—have interesting arithmetic properties. To facilitate the study of finite groups, it is convenient to introduce some terminology and notation. Definition Order of a Group

The number of elements of a group (finite or infinite) is called its order. We will use |G| to denote the order of G.

Thus, the group Z of integers under addition has infinite order, whereas the group U(10) 5 {1, 3, 7, 9} under multiplication modulo 10 has order 4. Definition Order of an Element

The order of an element g in a group G is the smallest positive integer n such that gn 5 e. (In additive notation, this would be ng 5 0.) If no such integer exists, we say that g has infinite order. The order of an element g is denoted by |g|.

So, to find the order of a group element g, you need only compute the sequence of products g, g2, g3, . . . , until you reach the identity for the first time. The exponent of this product (or coefficient if the operation is addition) is the order of g. If the identity never appears in the sequence, then g has infinite order. EXAMPLE 1 Consider U(15) 5 {1, 2, 4, 7, 8, 11, 13, 14} under multiplication modulo 15. This group has order 8. To find the order of 57

58

Groups

the element 7, say, we compute the sequence 71 5 7, 72 5 4, 73 5 13, 74 5 1, so |7| 5 4. To find the order of 11, we compute 111 5 11, 112 5 1, so |11| 5 2. Similar computations show that |1| 5 1, |2| 5 4, |4| 5 2, |8| 5 4, |13| 5 4, |14| 5 2. [Here is a trick that makes these calculations easier. Rather than compute the sequence 131, 132, 133, 134, we may observe that 13 5 22 mod 15, so that 132 5 (22)2 5 4, 133 5 22 ? 4 5 28, 134 5 (22)(28) 5 1.]† EXAMPLE 2 Consider Z10 under addition modulo 10. Since 1 ? 2 5 2, 2 ? 2 5 4, 3 ? 2 5 6, 4 ? 2 5 8, 5 ? 2 5 0, we know that |2| 5 5. Similar computations show that |0| 5 1, |7| 5 10, |5| 5 2, |6| 5 5. (Here 2 ? 2 is an abbreviation for 2 1 2, 3 ? 2 is an abbreviation for 2 1 2 1 2, etc.) EXAMPLE 3 Consider Z under ordinary addition. Here every nonzero element has infinite order, since the sequence a, 2a, 3a, . . . never includes 0 when a  0. The perceptive reader may have noticed among our examples of groups in Chapter 2 that some are subsets of others with the same binary operation. The group SL(2, R) in Example 17, for instance, is a subset of the group GL(2, R) in Example 9. Similarly, the group of complex numbers {1, 21, i, 2i} under multiplication is a subset of the group described in Example 14 for n equal to any multiple of 4. This situation arises so often that we introduce a special term to describe it. Definition Subgroup

If a subset H of a group G is itself a group under the operation of G, we say that H is a subgroup of G.

We use the notation H # G to mean that H is a subgroup of G. If we want to indicate that H is a subgroup of G but is not equal to G itself, we write H , G. Such a subgroup is called a proper subgroup. The subgroup {e} is called the trivial subgroup of G; a subgroup that is not {e} is called a nontrivial subgroup of G. Notice that Zn under addition modulo n is not a subgroup of Z under addition, since addition modulo n is not the operation of Z.

Subgroup Tests When determining whether or not a subset H of a group G is a subThe website www.google.com provides a convenient way to do modular arithmetic. For example, to compute 134 mod 15, just type 13ˆ4 mod 15 in the search box.

†

3 | Finite Groups; Subgroups

59

group of G, one need not directly verify the group axioms. The next three results provide simple tests that suffice to show that a subset of a group is a subgroup. Theorem 3.1 One-Step Subgroup Test Let G be a group and H a nonempty subset of G. If ab21 is in H whenever a and b are in H, then H is a subgroup of G. (In additive notation, if a 2 b is in H whenever a and b are in H, then H is a subgroup of G.)

PROOF Since the operation of H is the same as that of G, it is clear that this operation is associative. Next, we show that e is in H. Since H is nonempty, we may pick some x in H. Then, letting a 5 x and b 5 x in the hypothesis, we have e 5 xx21 5 ab21 is in H. To verify that x21 is in H whenever x is in H, all we need to do is to choose a 5 e and b 5 x in the statement of the theorem. Finally, the proof will be complete when we show that H is closed; that is, if x, y belong to H, we must show that xy is in H also. Well, we have already shown that y21 is in H whenever y is; so, letting a 5 x and b 5 y21, we have xy 5 x(y21)21 5 ab21 is in H. Although we have dubbed Theorem 3.1 the “One-Step Subgroup Test,’’ there are actually four steps involved in applying the theorem. (After you gain some experience, the first three steps will be routine.) Notice the similarity between the last three steps listed below and the three steps involved in the Principle of Mathematical Induction. 1. Identify the property P that distinguishes the elements of H; that is, identify a defining condition. 2. Prove that the identity has property P. (This verifies that H is nonempty.) 3. Assume that two elements a and b have property P. 4. Use the assumption that a and b have property P to show that ab21 has property P. The procedure is illustrated in Examples 4 and 5. EXAMPLE 4 Let G be an Abelian group with identity e. Then H 5 {x [ G | x2 5 e} is a subgroup of G. Here, the defining property of H is the condition x2 5 e. So, we first note that e2 5 e, so that H is nonempty. Now we assume that a and b belong to H. This means that a2 5 e and b2 5 e. Finally, we must show that (ab21)2 5 e. Since G is Abelian, (ab21)2 5 ab21ab21 5 a2(b21)2 5 a2(b2)21 5 ee21 5 e. Therefore, ab21 belongs to H and, by the One-Step Subgroup Test, H is a subgroup of G.

60

Groups

In many instances, a subgroup will consist of all elements that have a particular form. Then the property P is that the elements have that particular form. This is illustrated in the following example. EXAMPLE 5 Let G be an Abelian group under multiplication with identity e. Then H 5 {x2 | x [ G} is a subgroup of G. (In words, H is the set of all “squares.”) Since e2 5 e, the identity has the correct form. Next, we write two elements of H in the correct form, say, a2 and b2. We must show that a2(b2)21 also has the correct form; that is, a2(b2)21 is the square of some element. Since G is Abelian, we may write a2(b2)21 as (ab21)2, which is the correct form. Thus, H is a subgroup of G. Beginning students often prefer to use the next theorem instead of Theorem 3.1. Theorem 3.2 Two-Step Subgroup Test Let G be a group and let H be a nonempty subset of G. If ab is in H whenever a and b are in H (H is closed under the operation), and a21 is in H whenever a is in H (H is closed under taking inverses), then H is a subgroup of G.

PROOF By Theorem 3.1, it suffices to show that a, b [ H implies ab21 [ H. So, we suppose that a, b [ H. Since H is closed under taking inverses, we also have b21 [ H. Thus, ab21 [ H by closure under multiplication. When applying the “Two-Step Subgroup Test,” we proceed exactly as in the case of the “One-Step Subgroup Test,” except we use the assumption that a and b have property P to prove that ab has property P and that a21 has property P. How do you prove that a subset of a group is not a subgroup? Here are three possible ways, any one of which guarantees that the subset is not a subgroup: 1. Show that the identity is not in the set. 2. Exhibit an element of the set whose inverse is not in the set. 3. Exhibit two elements of the set whose product is not in the set. EXAMPLE 6 Let G be the group of nonzero real numbers under multiplication, H 5 {x [ G | x 5 1 or x is irrational} and K 5 {x [ G | x $ 1}. Then H is not a subgroup of G, since "2 [ H but "2 ? "2 5 2 o H. Also, K is not a subgroup, since 2 [ K but 221 o K.

3 | Finite Groups; Subgroups

61

When dealing with finite groups, it is easier to use the following subgroup test. Theorem 3.3 Finite Subgroup Test Let H be a nonempty finite subset of a group G. If H is closed under the operation of G, then H is a subgroup of G.

PROOF In view of Theorem 3.2, we need only prove that a21 [ H whenever a [ H. If a 5 e, then a21 5 a and we are done. If a  e, consider the sequence a, a2, . . . . By closure, all of these elements belong to H. Since H is finite, not all of these elements are distinct. Say ai 5 a j and i . j. Then, ai2j 5 e; and since a  e, i 2 j . 1. Thus, aai2j21 5 ai2j 5 e and, therefore, ai2j21 5 a21. But, i 2 j 2 1 $ 1 implies ai2j21 [ H and we are done.

Examples of Subgroups The proofs of the next few theorems show how our subgroup tests work. We first introduce an important notation. For any element a from a group, we let a denote the set {an | n [ Z}. In particular, observe that the exponents of a include all negative integers as well as 0 and the positive integers (a0 is defined to be the identity). Theorem 3.4 a Is a Subgroup Let G be a group, and let a be any element of G. Then, a is a subgroup of G.

PROOF Since a [ a, a is not empty. Let an, am [ a. Then, an(am)21 5 an2m [ a; so, by Theorem 3.1, a is a subgroup of G. The subgroup a is called the cyclic subgroup of G generated by a. In the case that G 5 a, we say that G is cyclic and a is a generator of G. (A cyclic group may have many generators.) Notice that although the list . . . , a22, a21, a0, a1, a2, . . . has infinitely many entries, the set {an | n [ Z} might have only finitely many elements. Also note that, since aia j 5 ai1j 5 a j1i 5 a j a i , every cyclic group is Abelian. 32

EXAMPLE 7 In U(10), 3 5 {3, 9, 7, 1} 5 U(10), for 31 5 3, 5 9, 33 5 7, 34 5 1, 35 5 34 ? 3 5 1 ? 3, 36 5 34 ? 32 5 9, . . . ; 321 5 7

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Groups

(since 3 ? 7 5 1), 322 5 9, 323 5 3, 324 5 1, 325 5 324 ? 321 5 1 ? 7, 326 5 324 ? 322 5 1 ? 9 5 9, . . . . EXAMPLE 8 In Z10, 2 5 {2, 4, 6, 8, 0}. Remember, an means na when the operation is addition. EXAMPLE 9 In Z, 21 5 Z. Here each entry in the list . . . , 22(21), 21(21), 0(21), 1(21), 2(21), . . . represents a distinct group element. EXAMPLE 10 In Dn, the dihedral group of order 2n, let R denote a rotation of 360/n degrees. Then, Rn 5 R360° 5 e,

Rn11 5 R,

Rn12 5 R2, . . . .

Similarly, R21 5 Rn21, R22 5 Rn22, . . . , so that R 5 {e, R, . . . , Rn21}. We see, then, that the powers of R “cycle back” periodically with period n. Visually, raising R to successive positive powers is the same as moving counterclockwise around the following circle one node at a time, whereas raising R to successive negative powers is the same as moving around the circle clockwise one node at a time. Rn 5 e Rn11 5 R

Rn12 5 R2

R21 5 Rn21

R22 5 Rn22

In Chapter 4 we will show that |a| 5 |a|; that is, the order of the subgroup generated by a is the order of a itself. (Actually, the definition of |a| was chosen to ensure the validity of this equation.) We next consider one of the most important subgroups. Definition Center of a Group

The center, Z(G), of a group G is the subset of elements in G that commute with every element of G. In symbols,

Z(G) 5 {a [ G | ax 5 xa for all x in G}. [The notation Z(G) comes from the fact that the German word for center is Zentrum. The term was coined by J. A. de Seguier in 1904.]

Theorem 3.5 Center Is a Subgroup The center of a group G is a subgroup of G.

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PROOF For variety, we shall use Theorem 3.2 to prove this result. Clearly, e [ Z(G), so Z(G) is nonempty. Now, suppose a, b [ Z(G). Then (ab)x 5 a(bx) 5 a(xb) 5 (ax)b 5 (xa)b 5 x(ab) for all x in G; and, therefore, ab [ Z(G). Next, assume that a [ Z(G). Then we have ax 5 xa for all x in G. What we want is a21x 5 xa21 for all x in G. Informally, all we need do to obtain the second equation from the first one is simultaneously to bring the a’s across the equals sign: ax 5 xa becomes xa21 5 a21x. (Be careful here; groups need not be commutative. The a on the left comes across as a21 on the left, and the a on the right comes across as a21 on the right.) Formally, the desired equation can be obtained from the original one by multiplying it on the left and right by a21, like so: a21(ax)a21 5 a21(xa)a21, (a21a)xa21 5 a21x(aa21), exa21 5 a21xe, xa21 5 a21x. This shows that a21 [ Z(G) whenever a is. For practice, let’s determine the centers of the dihedral groups. EXAMPLE 11 For n $ 3, Z(Dn ) 5 e

5R0, R1806 5R06

when n is even, when n is odd.

To verify this, first observe that since every rotation in Dn is a power of R360/n, rotations commute with rotations. We now investigate when a rotation commutes with a reflection. Let R be any rotation in Dn and let F be any reflection in Dn. Observe that since RF is a reflection we have RF 5 (RF )21 5 F21 R21 5 FR21. Thus it follows that R and F commute if and only if FR 5 RF 5 FR21. By cancellation, this holds if and only if R 5 R21. But R 5 R21 only when R 5 R0 or R 5 R180, and R180 is in Dn only when n is even. So, we have proved that Z(Dn) 5 {R0} when n is odd and Z(Dn) 5 {R0, R180} when n is even. Although an element from a non-Abelian group does not necessarily commute with every element of the group, there are always some elements with which it will commute. For example, every element a

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Groups

commutes with all powers of a. This observation prompts the next definition and theorem. Definition Centralizer of a in G

Let a be a fixed element of a group G. The centralizer of a in G, C(a), is the set of all elements in G that commute with a. In symbols, C(a) 5 {g [ G | ga 5 ag}.

EXAMPLE 12 In D4, we have the following centralizers: C(R0) 5 D4 5 C(R180), C(R90) 5 {R0, R90, R180, R270} 5 C(R270), C(H) 5 {R0, H, R180, V} 5 C(V), C(D) 5 {R0, D, R180, D9} 5 C(D9). Notice that each of the centralizers in Example 12 is actually a subgroup of D4. The next theorem shows that this was not a coincidence. Theorem 3.6 C(a) Is a Subgroup For each a in a group G, the centralizer of a is a subgroup of G.

PROOF A proof similar to that of Theorem 3.5 is left to the reader to supply (Exercise 25). Notice that for every element a of a group G, Z(G) # C(a). Also, observe that G is Abelian if and only if C(a) 5 G for all a in G.

Exercises The purpose of proof is to understand, not to verify. ARNOLD ROSS

1. For each group in the following list, find the order of the group and the order of each element in the group. What relation do you see between the orders of the elements of a group and the order of the group? Z12,

U(10),

U(12),

U(20),

D4

2. Let Q be the group of rational numbers under addition and let Q* be the group of nonzero rational numbers under multiplication. In Q, list the elements in 21 . In Q*, list the elements in 12 .

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65

3. Let Q and Q* be as in Exercise 2. Find the order of each element in Q and in Q*. 4. Prove that in any group, an element and its inverse have the same order. 5. Without actually computing the orders, explain why the two elements in each of the following pairs of elements from Z30 must have the same order: {2, 28}, {8, 22}. Do the same for the following pairs of elements from U(15): {2, 8}, {7, 13}. 6. Suppose that a is a group element and a6 5 e. What are the possibilities for |a|? Provide reasons for your answer. 7. If a is a group element and a has infinite order, prove that am ? an when m ? n. 8. Let x belong to a group. If x2 2 e and x6 5 e, prove that x4 2 e and x5  e. What can we say about the order of x? 9. Show that if a is an element of a group G, then |a| # |G|. 10. Show that U(14) 5 3 5 5. [Hence, U(14) is cyclic.] Is U(14) 5 11? 11. Show that U(20) 2 k for any k in U(20). [Hence, U(20) is not cyclic.] 12. Prove that an Abelian group with two elements of order 2 must have a subgroup of order 4. 13. Find groups that contain elements a and b such that |a| 5 |b| 5 2 and a. |ab| 5 3, b. |ab| 5 4, c. |ab| 5 5. Can you see any relationship among |a|, |b|, and |ab|? 14. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30, and 40. Determine H. 15. Suppose that H is a proper subgroup of Z under addition and that H contains 12, 30 and 54. What are the possibilities for H? 16. Prove that the dihedral group of order 6 does not have a subgroup of order 4. 17. For each divisor k . 1 of n, let Uk(n) 5 {x [ U(n) | x mod k 5 1}. [For example, U3(21) 5 {1, 4, 10, 13, 16, 19} and U7(21) 5 {1, 8}.] List the elements of U4(20), U5(20), U5(30), and U10(30). Prove that Uk(n) is a subgroup of U(n). Let H 5 {x [ U(10) | x mod 3 5 1}. Is H a subgroup of U(10)? (This exercise is referred to in Chapter 8.) 18. If H and K are subgroups of G, show that H > K is a subgroup of G. (Can you see that the same proof shows that the intersection of any number of subgroups of G, finite or infinite, is again a subgroup of G?)

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Groups

19. Let G be a group. Show that Z(G) 5 >a[GC(a). [This means the intersection of all subgroups of the form C(a).] 20. Let G be a group, and let a [ G. Prove that C(a) 5 C(a21). 21. For any group element a and any integer k, show that C(a) # C(a k). Use this fact to complete the following statement: “In a group, if R is an integer and x commutes with a, then . . . .” Is the converse true? 22. Complete the partial Cayley group table given below. 1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8

2 1 4 3 6 5 8 7

3 4 2 1 8 7 5 6

4 3 1 2 7 8 6 5

5 6 7 8 1

6 5 8 7

7 8 6 5

8 7 5 6

1 1 1

23. Suppose G is the group defined by the following Cayley table.

1 2 3 4 5 6 7 8

24. 25. 26.

27. 28. 29.

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7 8

2 1 4 3 6 5 8 7

3 8 5 2 7 4 1 6

4 7 6 1 8 3 2 5

5 6 7 8 1 2 3 4

6 5 8 7 2 1 4 3

7 4 1 6 3 8 5 2

8 3 2 5 4 7 6 1

a. Find the centralizer of each member of G. b. Find Z(G). c. Find the order of each element of G. How are these orders arithmetically related to the order of the group? If a and b are distinct group elements, prove that either a2 2 b2 or a3 2 b3. Prove Theorem 3.6. If H is a subgroup of G, then by the centralizer C(H) of H we mean the set {x [ G | xh 5 hx for all h [ H}. Prove that C(H) is a subgroup of G. Must the centralizer of an element of a group be Abelian? Must the center of a group be Abelian? Let G be an Abelian group with identity e and let n be some fixed integer. Prove that the set of all elements of G that satisfy the equation

3 | Finite Groups; Subgroups

30.

31. 32.

33. 34. 35.

67

xn 5 e is a subgroup of G. Give an example of a group G in which the set of all elements of G that satisfy the equation x2 5 e does not form a subgroup of G. (This exercise is referred to in Chapter 11.) Suppose a belongs to a group and |a| 5 5. Prove that C(a) 5 C(a3). Find an element a from some group such that |a| 5 6 and C(a)  C(a3). Determine all finite subgroups of R*, the group of nonzero real numbers under multiplication. Suppose n is an even positive integer and H is a subgroup of Zn. Prove that either every member of H is even or exactly half of the members of H are even. Suppose a group contains elements a and b such that |a| 5 4, |b| 5 2, and a3b 5 ba. Find |ab|. Suppose a and b are group elements such that |a| 5 2, b ? e, and aba 5 b 2. Determine |b|. Let a be a group element of order n, and suppose that d is a positive divisor of n. Prove that |ad | 5 n/d. 0 21 0 1 d and B 5 c d from 1 0 21 21 SL(2, R). Find |A|, |B|, and |AB|. Does your answer surprise you?

36. Consider the elements A 5 c 37. Consider the element A 5 c

1 1 d in SL(2, R). What is the order of 0 1

1 1 d as a member of SL(2, Zp) (p is a prime), 0 1 what is the order of A? 38. For any positive integer n and any angle u, show that in the group SL(2, R), A? If we view A 5 c

c

n

cos u 2 sin u cos nu d 5 c sin u cos u sin nu

Use this formula to find the order of c

cos 60 ° sin 60 °

(Geometrically, c

2 sin nu d. cos nu

2 sin 60° cos "2° d and c cos 60° sin "2°

2 sin "2° d. cos "2°

cos u 2 sin u d represents a rotation of the plane sin u cos u

u degrees.) 39. Let G be the symmetry group of a circle. Show that G has elements of every finite order as well as elements of infinite order.

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Groups

40. Let x belong to a group and |x| 5 6. Find |x2|, |x3|, |x4|, and |x5|. Let y belong to a group and |y| 5 9. Find |y i| for i 5 2, 3, . . . , 8. Do these examples suggest any relationship between the order of the power of an element and the order of the element? 41. D4 has seven cyclic subgroups. List them. Find a subgroup of D4 of order 4 that is not cyclic. 42. U(15) has six cyclic subgroups. List them. 43. Prove that a group of even order must have an element of order 2. 44. Suppose G is a group that has exactly eight elements of order 3. How many subgroups of order 3 does G have? 45. Let H be a subgroup of a finite group G. Suppose that g belongs to G and n is the smallest positive integer such that gn [ H. Prove that n divides |g|. 46. Compute the orders of the following groups. a. U(3), U(4), U(12) b. U(5), U(7), U(35) c. U(4), U(5), U(20) d. U(3), U(5), U(15) On the basis of your answers, make a conjecture about the relationship among |U(r)|, |U(s)|, and |U(rs)|. 47. Let R* be the group of nonzero real numbers under multiplication and let H 5 {x [ R* | x2 is rational}. Prove that H is a subgroup of R*. Can the exponent 2 be replaced by any positive integer and still have H be a subgroup? 48. Compute |U(4)|, |U(10)|, and |U(40)|. Do these groups provide a counterexample to your answer to Exercise 46? If so, revise your conjecture. 49. Find a cyclic subgroup of order 4 in U(40). 50. Find a noncyclic subgroup of order 4 in U(40). 51. Let G 5 e c

a b d 0 a, b, c, d c d

[Z f

under addition. Let H 5

a b d P G| a 1 b 1 c 1 d 5 0f. Prove that H is a subgroup of G. c d What if 0 is replaced by 1? 52. Let H 5 {A [ GL(2, R)| det A is an integer power of 2}. Show that H is a subgroup of GL(2, R). 53. Let H be a subgroup of R under addition. Let K 5 {2a | a [ H}. Prove that K is a subgroup of R* under multiplication. ec

3 | Finite Groups; Subgroups

69

54. Let G be a group of functions from R to R*, where the operation of G is multiplication of functions. Let H 5 { f [ G | f(2) 5 1}. Prove that H is a subgroup of G. Can 2 be replaced by any real number? a 0 55. Let G 5 GL(2, R) and H 5 e c d 0 a and b are nonzero inte0 b gers f under the operation of matrix multiplication. Prove or

56. 57.

58.

59.

60. 61.

62.

disprove that H is a subgroup of GL(2, R). Let H 5 {a 1 bi | a, b [ R, ab $ 0}. Prove or disprove that H is a subgroup of C under addition. Let H 5 {a 1 bi | a, b [ R, a2 1 b2 5 1}. Prove or disprove that H is a subgroup of C* under multiplication. Describe the elements of H geometrically. The smallest subgroup containing a collection of elements S is the subgroup H with the property that if K is any subgroup containing S then K also contains H. (So, the smallest subgroup containing S is contained in every subgroup that contains S.) The notation for this subgroup is S. In the group Z, find a. 8, 14 b. 8, 13 c. 6, 15 d. m, n e. 12, 18, 45. In each part, find an integer k such that the subgroup is k. Let G 5 GL(2, R). 1 1 a. Find C a c d b. 1 0 0 1 b. Find C a c d b. 1 0 c. Find Z(G). Let G be a finite group with more than one element. Show that G has an element of prime order. Let a belong to a group and |a| 5 m. If n is relatively prime to m, show that a can be written as the nth power of some element in the group. Let G be a finite Abelian group and let a and b belong to G. Prove that the set Ka, bL 5 {ai b j | i, j [ Z} is a subgroup of G. What can you say about |Ka, bL| in terms of |a| and |b|?

70

Groups

Computer Exercises A Programmer’s Lament I really hate this damned machine; I wish that they would sell it It never does quite what I want but only what I tell it. DENNIE L. VAN TASSEL, The Compleat Computer

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines the cyclic subgroups of U(n) generated by each k in U(n) (n , 100). Run the program for n 5 12, 15, and 30. Compare the order of the subgroups with the order of the group itself. What arithmetic relationship do these integers have? 2. The program lists the elements of Zn that generate all of Zn—that is, those elements k, 0 # k # n 2 1, for which Zn 5 k. How does this set compare with U(n)? Make a conjecture. 3. This software does the following: For each pair of elements a and b from U(n) (n , 100), it prints |a|, |b|, and |ab| on the same line. Run the program for several values of n. Is there an arithmetic relationship between |ab| and |a| and |b|? 4. This exercise repeats Exercise 3 for Zn using a 1 b in place of ab. 5. This software computes the order of elements in GL(2, Zp). Enter several choices for matrices A and B. The software returns |A|, |B|, |AB|, |BA|, |A21BA|, and |B21AB|. Do you see any relationship between |A|, |B| and |AB|? Do you see any relationship between |AB| and |BA|? Make a conjecture about this relationship. Test your conjecture for several other choices for A and B. Do you see any relationship between |B| and |A21BA|? Do you see any relationship between |A| and |B21AB|? Make a conjecture about this relationship. Test your conjecture for several other choices for A and B.

Suggested Readings Ruth Berger, “Hidden Group Structure,” Mathematics Magazine 78 (2005): 45–48.

3 | Finite Groups; Subgroups

71

In this note, the author investigates groups obtained from U(n) by multiplying each element by some k in U(n). Such groups have identities that are not obvious. J. Gallian and M. Reid, “Abelian Forcing Sets,” American Mathematical Monthly 100 (1993): 580–582. A set S is called Abelian forcing if the only groups that satisfy (ab)n 5 anbn for all a and b in the group and all n in S are the Abelian ones. This paper characterizes the Abelian forcing sets. Gina Kolata, “Perfect Shuffles and Their Relation to Math,” Science 216 (1982): 505–506. This is a delightful nontechnical article that discusses how group theory and computers were used to solve a difficult problem about shuffling a deck of cards. Serious work on the problem was begun by an undergraduate student as part of a programming course.

Suggested Software Allen Hibbard and Kenneth Levasseur, Exploring Abstract Algebra with Mathematica, New York: Springer-Verlag, 1999. This book, intended as a supplement for a course in abstract algebra, consists of 14 group labs, 13 ring labs, and documentation for the Abstract Algebra software on which the labs are based. The software uses the Mathematica language, and only a basic familiarity with the program is required. The software can be freely downloaded at http://www .central.edu/eaam/ and can be used independently of the book. This article can be downloaded at http://www.d.umn.edu/~jgallian/forcing.pdf

4 Cyclic Groups The notion of a “group,” viewed only 30 years ago as the epitome of sophistication, is today one of the mathematical concepts most widely used in physics, chemistry, biochemistry, and mathematics itself. ALEXEY SOSINSKY, 1991

Properties of Cyclic Groups Recall from Chapter 3 that a group G is called cyclic if there is an element a in G such that G 5 {an | n [ Z}. Such an element a is called a generator of G. In view of the notation introduced in the preceding chapter, we may indicate that G is a cyclic group generated by a by writing G 5 a. In this chapter, we examine cyclic groups in detail and determine their important characteristics. We begin with a few examples. EXAMPLE 1 The set of integers Z under ordinary addition is cyclic. Both 1 and 21 are generators. (Recall that, when the operation is addition, 1n is interpreted as 1 1 1 1 ??? 1 1 n terms

when n is positive and as (21) 1 (21) 1 ? ? ? 1 (21) |n| terms

when n is negative.) EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} for n $ 1 is a cyclic group under addition modulo n. Again, 1 and 21 5 n 2 1 are generators.

72

4 | Cyclic Groups

73

Unlike Z, which has only two generators, Zn may have many generators (depending on which n we are given). EXAMPLE 3 Z8 5 k1l 5 k3l 5 k5l 5 k7l. To verify, for instance, that Z8 5 k3l, we note that k3l 5 {3, 3 1 3, 3 1 3 1 3, . . .} is the set {3, 6, 1, 4, 7, 2, 5, 0} 5 Z8. Thus, 3 is a generator of Z8. On the other hand, 2 is not a generator, since k2l 5 {0, 2, 4, 6} 2 Z8. EXAMPLE 4 (See Example 11 in Chapter 2.) U(10) 5 {1, 3, 7, 9} 5 {30, 31, 33, 32} 5 3. Also, {1, 3, 7, 9} 5 {70, 73, 71, 72} 5 7. So both 3 and 7 are generators for U(10). Quite often in mathematics, a “nonexample” is as helpful in understanding a concept as an example. With regard to cyclic groups, U(8) serves this purpose; that is, U(8) is not a cyclic group. How can we verify this? Well, note that U(8) 5 {1, 3, 5, 7}. But 1 5 {1} 3 5 {3, 1} 5 5 {5, 1} 7 5 {7, 1} so U(8) 2 a for any a in U(8). With these examples under our belts, we are now ready to tackle cyclic groups in an abstract way and state their key properties. Theorem 4.1 Criterion for ai 5 a j Let G be a group, and let a belong to G. If a has infinite order, then ai 5 a j if and only if i 5 j. If a has finite order, say, n, then a 5 {e, a, a2, . . . , an–1} and ai 5 a j if and only if n divides i – j.

PROOF If a has infinite order, there is no nonzero n such that an is the identity. Since ai 5 a j implies ai2j 5 e, we must have i 2 j 5 0, and the first statement of the theorem is proved. Now assume that |a| 5 n. We will prove that a 5 {e, a, . . . , an21}. Certainly, the elements e, a, . . . , an21 are in a. Now, suppose that ak is an arbitrary member of a. By the division algorithm, there exist integers q and r such that k 5 qn 1 r with 0 # r , n.

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Groups

Then ak 5 aqn1r 5 aqnar 5 (an)qar 5 ear 5 ar, so that ak [ {e, a, a2, . . . , an21}. This proves that a 5 {e, a, a2, . . . , an21}. Next, we assume that ai 5 a j and prove that n divides i 2 j. We begin by observing that ai 5 a j implies ai2j 5 e. Again, by the division algorithm, there are integers q and r such that i 2 j 5 qn 1 r

with

0 # r , n.

Then ai2j 5 aqn1r, and therefore e 5 ai2j 5 aqn1r 5 (an)qar 5 eqar 5 ear 5 ar. Since n is the least positive integer such that an is the identity, we must have r 5 0, so that n divides i 2 j. Conversely, if i 2 j 5 nq, then ai2j 5 anq 5 eq 5 e, so that i a 5 a j. Theorem 4.1 reveals the reason for the dual use of the notation and terminology for the order of an element and the order of a group. Corollary 1 |a| 5 |a| For any group element a, |a| 5 |a|.

One special case of Theorem 4.1 occurs so often that it deserves singling out. Corollary 2 a k 5 e Implies That |a| Divides k Let G be a group and let a be an element of order n in G. If a k 5 e, then n divides k.

PROOF Since ak 5 e 5 a0, we know by Theorem 4.1 that n divides k 2 0. Theorem 4.1 and its corollaries for the case |a| 5 6 are illustrated in Figure 4.1. What is important about Theorem 4.1 in the finite case is that it says that multiplication in a is essentially done by addition modulo n. That is, if (i 1 j) mod n 5 k, then aia j 5 ak. Thus, no matter what group G is, or how the element a is chosen, multiplication in a works the same as addition in Zn whenever |a| 5 n. Similarly, if a has infinite order,

4 | Cyclic Groups

75

... a –6 = a 0 = a 6 ... ... a – 5 = a = a 7...

... a –1 = a 5 = a 11...

... a – 4 = a 2 = a 8 ...

... a –2 = a 4 = a 10... ... a –3 = a 3 = a 9...

Figure 4.1

then multiplication in a works the same as addition in Z, since aia j 5 ai1j and no modular arithmetic is done. For these reasons, the cyclic groups Zn and Z serve as prototypes for all cyclic groups, and algebraists say that there is essentially only one cyclic group of each order. What is meant by this is that, although there may be many different sets of the form {an | n [ Z}, there is essentially only one way to operate on these sets. Algebraists do not really care what the elements of a set are; they care only about the algebraic properties of the set—that is, the ways in which the elements of a set can be combined. We will return to this theme in the chapter on isomorphisms (Chapter 6). The next theorem provides a simple method for computing |ak| knowing only |a|, and its first corollary provides a simple way to tell when ai 5 a j . Theorem 4.2 ak 5 agcd(n,k) Let a be an element of order n in a group and let k be a positive integer. Then ak 5 agcd(n,k) and |ak| 5 n/gcd(n,k).

PROOF To simplify the notation, let d 5 gcd(n,k) and let k 5 dr. Since ak 5 (ad)r, we have by closure that ak # ad. By Theorem 0.2 (the gcd theorem), there are integers s and t such that d 5 ns 1 kt. So, ad 5 ans1kt 5 ansakt 5 (an)s(ak)t 5 e(ak)t 5 (ak)t [ ak. This proves ad # ak. So, we have verified that ak 5 agcd(n,k). We prove the second part of the theorem by showing first that |ad| 5 n/d for any divisor d of n. Clearly, (ad)n/d 5 an 5 e, so that |ad| # n/d. On the other hand, if i is a positive integer less than n/d, then (ad)i 2 e by definition of |a|. We now apply this fact with d 5 gcd(n,k) to obtain |ak| 5 |ak| 5 |agcd(n,k)| 5 |agcd(n,k)| 5 n/gcd(n,k). The advantage of Theorem 4.2 is that it allows us to replace one generator of a cyclic subgroup with a more convenient one. For example,

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Groups

if |a| 5 30, we have a26 5 a2, a23 5 a, a22 5 a2, a21 5 a3. From this we can easily see that |a23| 5 30 and |a22| 5 15. Moreover, if one wants to list the elements of, say, a21, it is easier to list the elements of a3 instead. (Try it doing it both ways!). Theorem 4.2 establishes an important relationship between the order of an element in a finite cyclic group and the order of the group. Corollary 1 Orders of Elements in Finite Cyclic Groups In a finite cyclic group, the order of an element divides the order of the group.

Corollary 2 Criterion for ai   a j  and |ai |  |a j | Let |a| 5 n. Then ai 5 a j if and only if gcd(n, i) 5 gcd(n, j) and |ai| 5 |a j| if and only if gcd(n, i) 5 gcd(n, j) .

PROOF Theorem 4.2 shows that ai 5 agcd(n,i) and a j 5 agcd(n,j), so that the proof reduces to proving that agcd(n,i) 5 agcd(n,j) if and only if gcd(n, i) 5 gcd(n, j). Certainly, gcd(n, i) 5 gcd(n, j) implies that agcd(n,i) 5 agcd(n, j). On the other hand, agcd(n,i) 5 agcd(n,j) implies that |agcd(n,i)|5 |agcd(n,j)|, so that by the second conclusion of Theorem 4.2, we have n/gcd(n, i) 5 n/gcd(n, j), and therefore gcd(n, i) 5 gcd(n, j). The second part of the corollary follows from the first part and Corollary 1 of Theorem 4.1. The next two corollaries are important special cases of the preceding corollary. Corollary 3 Generators of Finite Cyclic Groups Let |a| 5 n. Then a 5 a j if and only if gcd(n, j) 5 1 and |a| 5 |a j| if and only if gcd(n, j) 5 1.

Corollary 4 Generators of Zn An integer k in Zn is a generator of Zn if and only if gcd(n, k) 5 1.

The value of Corollary 3 is that once one generator of a cyclic group has been found, all generators of the cyclic group can easily be determined.

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For example, consider the subgroup of all rotations in D6. Clearly, one generator is R60. And, since |R60| 5 6, we see by Corollary 3 that the only other generator is (R60)5 5 R300. Of course, we could have readily deduced this information without the aid of Corollary 3 by direct calculations. So, to illustrate the real power of Corollary 3, let us use it to find all generators of the cyclic group U(50). First, note that direct computations show that |U(50)| 5 20 and that 3 is one of its generators. Thus, in view of Corollary 3, the complete list of generators for U(50) is 3 mod 50 5 3, 33 mod 50 5 27, 37 mod 50 5 37, 39 mod 50 5 33,

311 mod 50 5 47, 313 mod 50 5 23, 317 mod 50 5 13, 319 mod 50 5 17.

Admittedly, we had to do some arithmetic here, but it certainly entailed much less work than finding all the generators by simply determining the order of each element of U(50) one by one. The reader should keep in mind that Theorem 4.2 and its corollaries apply only to elements of finite order.

Classification of Subgroups of Cyclic Groups The next theorem tells us how many subgroups a finite cyclic group has and how to find them. Theorem 4.3 Fundamental Theorem of Cyclic Groups Every subgroup of a cyclic group is cyclic. Moreover, if |a| 5 n, then the order of any subgroup of a is a divisor of n; and, for each positive divisor k of n, the group a has exactly one subgroup of order k—namely, an/ k .

Before we prove this theorem, let’s see what it means. Understanding what a theorem means is a prerequisite to understanding its proof. Suppose G 5 a and G has order 30. The first and second parts of the theorem say that if H is any subgroup of G, then H has the form a30/k for some k that is a divisor of 30. The third part of the theorem says that G has one subgroup of each of the orders 1, 2, 3, 5, 6, 10, 15, and 30—and no others. The proof will also show how to find these subgroups. PROOF Let G 5 a and suppose that H is a subgroup of G. We must show that H is cyclic. If it consists of the identity alone, then clearly H is cyclic. So we may assume that H 2 {e}. We now claim that H contains

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an element of the form at, where t is positive. Since G 5 a, every element of H has the form at; and when at belongs to H with t , 0, then a2t belongs to H also and 2t is positive. Thus, our claim is verified. Now let m be the least positive integer such that am [ H. By closure, am # H. We next claim that H 5 am. To prove this claim, it suffices to let b be an arbitrary member of H and show that b is in am. Since b [ G 5 a, we have b 5 ak for some k. Now, apply the division algorithm to k and m to obtain integers q and r such that k 5 mq 1 r where 0 # r , m. Then ak 5 amq1r 5 amqar, so that ar 5 a2mqak. Since ak 5 b [ H and a2mq 5 (am)2q is in H also, ar [ H. But, m is the least positive integer such that am [ H, and 0 # r , m, so r must be 0. Therefore, b 5 ak 5 amq 5 (am)q [ am. This proves the assertion of the theorem that every subgroup of a cyclic group is cyclic. To prove the next portion of the theorem, suppose that |a| 5 n and H is any subgroup of a. We have already shown that H 5 am, where m is the least positive integer such that am [ H. Using e 5 b 5 an as in the preceding paragraph, we have n 5 mq. Finally, let k be any positive divisor of n. We will show that an/k is the one and only subgroup of a of order k. From Theorem 4.2, we see that an/k has order n/gcd(n, n/k) 5 n/(n/k) 5 k. Now let H be any subgroup of a of order k. We have already shown above that H 5 am, where m is a divisor of n. Then m 5 gcd(n, m) and k 5 |am| 5 |agcd(n,m)| 5 n/gcd (n, m) 5 n/m. Thus, m 5 n/k and H 5 an/k. Returning for a moment to our discussion of the cyclic group a, where a has order 30, we may conclude from Theorem 4.3 that the subgroups of a are precisely those of the form am, where m is a divisor of 30. Moreover, if k is a divisor of 30, the subgroup of order k is a30/k. So the list of subgroups of a is: a 5 {e, a, a2, . . . , a29} a2 5 {e, a2, a4, . . . , a28} a3 5 {e, a3, a6, . . . , a27} a5 5 {e, a5, a10, a15, a20, a25} a6 5 {e, a6, a12, a18, a24} a10 5 {e, a10, a20} a15 5 {e, a15} a30 5 {e}

order 30, order 15, order 10, order 6, order 5, order 3, order 2, order 1.

In general, if a has order n and k divides n, then an/k is the unique subgroup of order k. Taking the group in Theorem 4.3 to be Zn and a to be 1, we obtain the following important special case.

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Corollary Subgroups of Zn For each positive divisor k of n, the set n/k is the unique subgroup of Zn of order k; moreover, these are the only subgroups of Zn.

EXAMPLE 5 The list of subgroups of Z30 is 1 5 {0, 1, 2, . . . , 29} 2 5 {0, 2, 4, . . . , 28} 3 5 {0, 3, 6, . . . , 27} 5 5 {0, 5, 10, 15, 20, 25} 6 5 {0, 6, 12, 18, 24} 10 5 {0, 10, 20} 15 5 {0, 15} 30 5 {0}

order 30, order 15, order 10, order 6, order 5, order 3, order 2, order 1.

By combining Theorems 4.2 and 4.3, we can easily count the number of elements of each order in a finite cyclic group. For convenience, we introduce an important number-theoretic function called the Euler phi function. Let f(1) 5 1, and for any integer n . 1, let f(n) denote the number of positive integers less than n and relatively prime to n. Notice that by definition of the group U(n), |U(n)| 5 f(n). The first 12 values of f(n) are given in Table 4.1. Table 4.1 Values of f(n) 1

2 3 4

5 6

7

8

9 10

11 12

f(n) 1

1 2 2

4 2

6

4

6

10

n

4

4

Theorem 4.4 Number of Elements of Each Order in a Cyclic Group If d is a positive divisor of n, the number of elements of order d in a cyclic group of order n is f(d).

PROOF By Theorem 4.3, the group has exactly one subgroup of order d—call it a. Then every element of order d also generates the subgroup a and, by Corollary 3 of Theorem 4.2, an element a k generates a if and only if gcd(k, d) 5 1. The number of such elements is precisely f(d).

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Notice that for a finite cyclic group of order n, the number of elements of order d for any divisor d of n depends only on d. Thus, Z8, Z640, and Z80000 each have f(8) 5 4 elements of order 8. Although there is no formula for the number of elements of each order for arbitrary finite groups, we still can say something important in this regard. Corollary Number of Elements of Order d in a Finite Group In a finite group, the number of elements of order d is divisible by f(d).

PROOF If a finite group has no elements of order d, the statement is true, since f(d) divides 0. Now suppose that a [ G and |a| 5 d. By Theorem 4.4, we know that a has f(d) elements of order d. If all elements of order d in G are in a, we are done. So, suppose that there is an element b in G of order d that is not in a. Then, b also has f(d) elements of order d. This means that we have found 2f(d) elements of order d in G provided that a and b have no elements of order d in common. If there is an element c of order d that belongs to both a and b, then we have a 5 c 5 b, so that b [ a, which is a contradiction. Continuing in this fashion, we see that the number of elements of order d in a finite group is a multiple of f(d). On its face, the value of Theorem 4.4 and its corollary seem limited for large values of n because it is tedious to determine the number of positive integers less than or equal to n and relatively prime to n by examining them one by one. However, the following properties of the f function make computing f(n) simple: For any prime p, f(pn ) 5 pn 2 pn21 (see Exercise 71) and for relatively prime m and n, f(mn) 5 f(m)f(n). Thus, f(40) 5 f(8)f(5) 5 4 ? 4 5 16; f(75) 5 f(52 )f(3) 5 (25 2 5) ? 2 5 40. The relationships among the various subgroups of a group can be illustrated with a subgroup lattice of the group. This is a diagram that includes all the subgroups of the group and connects a subgroup H at one level to a subgroup K at a higher level with a sequence of line segments if and only if H is a proper subgroup of K. Although there are many ways to draw such a diagram, the connections between the subgroups must be the same. Typically one attempts to present the diagram in an eye-pleasing fashion. The lattice diagram for Z30 is shown in Figure 4.2. Notice that 10 is a subgroup of both 2 and 5, but 6 is not a subgroup of 10.

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<1>

<5>

< 2> < 3> <10>

<15>

<6>

<0>

Figure 4.2 Subgroup lattice of Z30.

The precision of Theorem 4.3 can be appreciated by comparing the ease with which we are able to identify the subgroups of Z30 with that of doing the same for, say, U(30) or D30. And these groups have relatively simple structures among noncyclic groups. We will prove in Chapter 7 that a certain portion of Theorem 4.3 extends to arbitrary finite groups; namely, the order of a subgroup divides the order of the group itself. We will also see, however, that a finite group need not have exactly one subgroup corresponding to each divisor of the order of the group. For some divisors, there may be none at all, whereas for other divisors, there may be many. Indeed, D4, the dihedral group of order 8, has five subgroups of order 2 and three of order 4. One final remark about the importance of cyclic groups is appropriate. Although cyclic groups constitute a very narrow class of finite groups, we will see in Chapter 11 that they play the role of building blocks for all finite Abelian groups in much the same way that primes are the building blocks for the integers and that chemical elements are the building blocks for the chemical compounds.

Exercises It is not unreasonable to use the hypothesis. ARNOLD ROSS

1. Find all generators of Z6, Z8, and Z20. 2. Suppose that a, b, and c are cyclic groups of orders 6, 8, and 20, respectively. Find all generators of a, b, and c. 3. List the elements of the subgroups 20 and 10 in Z30 . Let a be a group element of order 30. List the elements of the subgroups a20  and a10 .

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4. List the elements of the subgroups 3 and 15 in Z18. Let a be a group element of order 18. List the elements of the subgroups a3 and a15. 5. List the elements of the subgroups 3 and 7 in U(20). 6. What do Exercises 3, 4, and 5 have in common? Try to make a generalization that includes these three cases. 7. Find an example of a noncyclic group, all of whose proper subgroups are cyclic. 8. Let a be an element of a group and let |a| 5 15. Compute the orders of the following elements of G. a. a3, a6, a9, a12 b. a5, a10 c. a2, a4, a8, a14 9. How many subgroups does Z20 have? List a generator for each of these subgroups. Suppose that G 5 a and |a| 5 20. How many subgroups does G have? List a generator for each of these subgroups. 10. In Z24 list all generators for the subgroup of order 8. Let G 5 a and let |a| 5 24. List all generators for the subgroup of order 8. 11. Let G be a group and let a [ G. Prove that a21 5 a. 12. In Z find all generators of the subgroup 3. If a has infinite order, find all generators of the subgroup a3 . 13. In Z24 find a generator for 21 > 10. Suppose that |a| 5 24. Find a generator for a21 > a10. In general, what is a generator for the subgroup am > an? 14. Suppose that a cyclic group G has exactly three subgroups: G itself, {e}, and a subgroup of order 7. What is |G|? What can you say if 7 is replaced with p where p is a prime? 15. Let G be an Abelian group and let H 5 {g [ G| |g| divides 12}. Prove that H is a subgroup of G. Is there anything special about 12 here? Would your proof be valid if 12 were replaced by some other positive integer? State the general result. 16. Find a collection of distinct subgroups a1, a2, . . . , an of Z240 with the property that a1 , a2 , ? ? ? , an with n as large as possible. 17. Complete the following statement: |a| 5 |a2| if and only if |a| . . . . 18. If a cyclic group has an element of infinite order, how many elements of finite order does it have? 19. List the cyclic subgroups of U(30).

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20. Suppose that G is an Abelian group of order 35 and every element of G satisfies the equation x35 5 e. Prove that G is cyclic. Does your argument work if 35 is replaced with 33? 21. Let G be a group and let a be an element of G. a. If a12 5 e, what can we say about the order of a? b. If am 5 e, what can we say about the order of a? c. Suppose that |G| 5 24 and that G is cyclic. If a8 2 e and a12 2 e, show that a 5 G. 22. Prove that a group of order 3 must be cyclic. 23. Let Z denote the group of integers under addition. Is every subgroup of Z cyclic? Why? Describe all the subgroups of Z. Let a be a group element with infinite order. Describe all subgroups of a. 24. For any element a in any group G, prove that a is a subgroup of C(a) (the centralizer of a). 25. If d is a positive integer, d 2 2, and d divides n, show that the number of elements of order d in Dn is f(d). How many elements of order 2 does Dn have? 26. Find all generators of Z. Let a be a group element that has infinite order. Find all generators of a. 27. Prove that C*, the group of nonzero complex numbers under multiplication, has a cyclic subgroup of order n for every positive integer n. 28. Let a be a group element that has infinite order. Prove that ai 5 a j  if and only if i 5 j. 29. List all the elements of order 8 in Z8000000. How do you know your list is complete? Let a be a group element such that |a| 5 8000000. List all elements of order 8 in a. How do you know your list is complete? 30. Suppose a and b belong to a group, a has odd order, and aba21 5 b21. Show that b2 5 e. 31. Let G be a finite group. Show that there exists a fixed positive integer n such that an 5 e for all a in G. (Note that n is independent of a.) 32. Determine the subgroup lattice for Z12. 33. Determine the subgroup lattice for Zp2q, where p and q are distinct primes. 34. Determine the subgroup lattice for Z8. 35. Determine the subgroup lattice for Zpn, where p is a prime and n is some positive integer. 36. Prove that a finite group is the union of proper subgroups if and only if the group is not cyclic. 37. Show that the group of positive rational numbers under multiplication is not cyclic.

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38. Consider the set {4, 8, 12, 16}. Show that this set is a group under multiplication modulo 20 by constructing its Cayley table. What is the identity element? Is the group cyclic? If so, find all of its generators. 39. Give an example of a group that has exactly 6 subgroups (including the trivial subgroup and the group itself). Generalize to exactly n subgroups for any positive integer n. 40. Let m and n be elements of the group Z. Find a generator for the group m > n. 41. Suppose that a and b are group elements that commute and have orders m and n. If a > b 5 {e}, prove that the group contains an element whose order is the least common multiple of m and n. Show that this need not be true if a and b do not commute. 42. Prove that an infinite group must have an infinite number of subgroups. 43. Let p be a prime. If a group has more than p 2 1 elements of order p, why can’t the group be cyclic? 44. Suppose that G is a cyclic group and that 6 divides |G|. How many elements of order 6 does G have? If 8 divides |G|, how many elements of order 8 does G have? If a is one element of order 8, list the other elements of order 8. 45. List all the elements of Z40 that have order 10. Let |x| 5 40. List all the elements of x that have order 10. 46. Reformulate the corollary of Theorem 4.4 to include the case when the group has infinite order. 47. Determine the orders of the elements of D33 and how many there are of each. 48. If G is a cyclic group and 15 divides the order of G, determine the number of solutions in G of the equation x15 5 e. If 20 divides the order of G, determine the number of solutions of x20 5 e. Generalize. 49. If G is an Abelian group and contains cyclic subgroups of orders 4 and 5, what other sizes of cyclic subgroups must G contain? Generalize. 50. If G is an Abelian group and contains cyclic subgroups of orders 4 and 6, what other sizes of cyclic subgroups must G contain? Generalize. 51. Prove that no group can have exactly two elements of order 2. 52. Given the fact that U(49) is cyclic and has 42 elements, deduce the number of generators that U(49) has without actually finding any of the generators.

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53. Let a and b be elements of a group. If |a| 5 10 and |b| 5 21, show that a > b 5 {e}. 54. Let a and b belong to a group. If |a| and |b| are relatively prime, show that a > b 5 {e}. 55. Let a and b belong to a group. If |a| 5 24 and |b| 5 10, what are the possibilities for |a > b|? 56. Prove that U(2n) (n $ 3) is not cyclic. 57. Suppose that a group G has at least nine elements x such that x8 5 e. Can you conclude that G is not cyclic? What if G has at least five elements x such that x4 5 e? Generalize. 58. Prove that Zn has an even number of generators if n . 2. What does this tell you about f(n)? 59. If |a5| 5 12, what are the possibilities for |a|? If |a4| 5 12, what are the possibilities for |a|? 60. Suppose that |x| 5 n. Find a necessary and sufficient condition on r and s such that x r  # x s . 61. Suppose a is a group element such that |a28| 5 10 and |a22| 5 20. Determine |a| . 62. Let a be group element such that |a| 5 48. For each part find a divisor k of 48 such that a. a21  5 ak  b. a14  5 ak  c. a18  5 ak . 63. Let p be a prime. Show that in a cyclic group of order pn 21, every element is a pth power (that is, every element can be written in the form ap for some a). 1 n d 0 n [ Z f is a cyclic subgroup of 64. Prove that H 5 e c 0 1 GL(2, R). 65. Let a and b belong to a group. If |a| 5 12, |b| 5 22, and a > b 2 {e}, prove that a6 5 b11. 66. Suppose that G is a finite group with the property that every nonidentity element has prime order (for example, D3 and D5). If Z(G) is not trivial, prove that every nonidentity element of G has the same order. 67. Let G be the set of all polynomials of the form ax 2 1 bx 1 c with coefficients from the set {0, 1, 2}. We can make G a group under addition by adding the polynomials in the usual way, except that we use modulo 3 to combine the coefficients. With this operation, prove that G is a group of order 27 that is not cyclic.

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68. Let r1 and r2 be rational numbers. Prove that the group G 5 5n1r1 1 n2r2 |n1 and n2 are integers6 under addition is cyclic. Generalize to the case where you have r1, r2 , . . . , rk rationals. 69. Let a and b belong to some group. Suppose that |a| 5 m and |b| 5 n and m and n are relatively prime. If ak 5 bk for some integer k, prove that mn divides k. 70. For every integer n greater than 2, prove that the group U(n2 2 1) is not cyclic. 71. Prove that for any prime p and positive integer n, f(pn ) 5 pn 2 pn21. 72. Give an example of an infinite group that has exactly two elements of order 4.

Computer Exercises The nerds are running the world now. JOE PISCOPO

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines if U(n) is cyclic. Run the program for n 5 8, 32, 64, and 128. Make a conjecture. Run the program for n 5 3, 9, 27, 81, 243, 5, 25, 125, 7, 49, 11, and 121. Make a conjecture. Run the program for n 5 12, 20, 28, 44, 52, 15, 21, 33, 39, 51, 57, 69, 35, 55, 65, and 85. Make a conjecture. 2. For any pair of positive integers m and n, let Zm % Zn 5 {(a, b) | a [ Zm, b [ Zn}. For any pair of elements (a, b) and (c, d) in Zm % Zn, define (a, b) 1 (c, d) 5 ((a 1 c) mod m, (b 1 d) mod n). [For example, in Z3 % Z4, we have (1, 2) 1 (2, 3) 5 (0, 1).] This software checks whether or not Zm % Zn is cyclic. Run the program for the following choices of m and n: (2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), and (4, 6). On the basis of this output, guess how m and n must be related for Zm % Zn to be cyclic. 3. In this exercise, a, b [ U(n). Define a, b 5 {aib j | 0 # i , |a|, 0 # j , |b|}. This software computes the orders of a, b, a, b, and a > b. Run the program for the following choices of a, b, and n: (21, 101, 550), (21, 49, 550), (7, 11, 100), (21, 31, 100), and

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(63, 77, 100). On the basis of your output, make a conjecture about arithmetic relationships among |a, b|, |a|, |b|, and |a > b|. 4. For each positive integer n, this software gives the order of U(n) and the order of each element in U(n). Do you see any relationship between the order of U(n) and the order of its elements? Run the program for n 5 8, 16, 32, 64, and 128. Make a conjecture about the number of elements of order 2 in U(2k) when k is at least 3. Make a conjecture about the number of elements of order 4 in U(2k) when k is at least 4. Make a conjecture about the number of elements of order 8 in U(2k) when k is at least 5. Make a conjecture about the maximum order of any element in U(2k) when k is at least 3. Try to find a formula for an element of order 4 in U(2k) when k is at least 4. 5. For each positive integer n, this software lists the number of elements of U(n) of each order. For each order d of some element of U(n), this software lists f(d) and the number of elements of order d. (Recall that f(d) is the number of positive integers less than or equal to d and relatively prime to d). Do you see any relationship between the number of elements of order d and f(d)? Run the program for n 5 3, 9, 27, 81, 5, 25, 125, 7, 49, and 343. Make a conjecture about the number of elements of order d and f(d) when n is a power of an odd prime. Run the program for n 5 6, 18, 54, 162, 10, 50, 250, 14, 98, and 686. Make a conjecture about the number of elements of order d and f(d) when n is twice a power of an odd prime. Make a conjecture about the number of elements of various orders in U( pk) and U(2pk) where p is an odd prime. 6. For each positive integer n, this software gives the order of U(n). Run the program for n 5 9, 27, 81, and 243. Try to guess a formula for the order of U(3k) when k is at least 2. Run the program for n 5 18, 54, 162, and 486. How does the order of U(2 ? 3k) appear to be related to the order of U(3k)? Run the program for n 5 25, 125, and 625. Try to guess a formula for the order of U(5k) when k is at least 2. Run the program for n 5 50, 250, and 1250. How does the order of U(2 ? 5k) appear to be related to the order of U(5k)? Run the program for n 5 49 and 343. Try to guess a formula for the order of U(7k) when k is at least 2. Run the program for n 5 98 and 686. How does the order of U(2 ? 7k) appear to be related to the order of U(7k)? Based on your guesses for U(3k), U(5k), and U(7k), guess a formula for the order of U( pk) when p is an odd prime and k is at least 2. What about the order of U(2pk) when p is an odd prime and k is at least 2. Does your formula also work when k is 1?

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Suggested Reading Deborah L. Massari, “The Probability of Regenerating a Cyclic Group,” Pi Mu Epsilon Journal 7 (1979): 3–6. In this easy-to-read paper, it is shown that the probability of a randomly chosen element from a cyclic group being a generator of the group depends only on the set of prime divisors of the order of the group, and not on the order itself. This article, written by an undergraduate student, received first prize in a Pi Mu Epsilon Paper Contest.

J. J. Sylvester I really love my subject. J. J. SYLVESTER

FPO

JAMES JOSEPH SYLVESTER was the most influential mathematician in America in the 19th century. Sylvester was born on September 3, 1814, in London and showed his mathematical genius early. At the age of 14, he studied under De Morgan and won several prizes for his mathematics, and at the unusually young age of 25, he was elected a Fellow of the Royal Society. After receiving B.A. and M.A. degrees from Trinity College in Dublin in 1841, Sylvester began a professional life that was to include academics, law, and actuarial careers. In 1876, at the age of 62, he was appointed to a prestigious position at the newly founded Johns Hopkins University. During his seven years at Johns Hopkins, Sylvester pursued research in pure mathematics with tremendous vigor and enthusiasm. He also founded the American Journal of Mathematics, the first journal in America devoted to mathematical research. Sylvester returned to England in 1884 to a professorship at Oxford, a position he held until his death on March 15, 1897. Sylvester’s major contributions to mathematics were in the theory of equations, matrix theory, determinant theory, and invariant theory (which he founded with Cayley). His writings and lectures—flowery and eloquent, pervaded with poetic flights, emotional expressions, bizarre utterances, and paradoxes—reflected the personality of this sensitive, excitable, and enthusiastic

†F.

man. We quote three of his students.† E. W. Davis commented on Sylvester’s teaching methods. Sylvester’s methods! He had none. “Three lectures will be delivered on a New Universal Algebra,” he would say; then, “The course must be extended to twelve.” It did last all the rest of that year. The following year the course was to be Substitutions-Theorie, by Netto. We all got the text. He lectured about three times, following the text closely and stopping sharp at the end of the hour. Then he began to think about matrices again. “I must give one lecture a week on those,” he said. He could not confine himself to the hour, nor to the one lecture a week. Two weeks were passed, and Netto was forgotten entirely and never mentioned again. Statements like the following were not infrequent in his lectures: “I haven’t proved this, but I am as sure as I can be of anything that it must be so. From this it will follow, etc.” At the next lecture it turned out that what he was so sure of was false. Never mind, he kept on forever guessing and trying, and presently a wonderful discovery followed, then another and another. Afterward he would go back and work it all over again, and surprise us with all sorts of side lights. He then made another leap in the dark, more treasures were discovered, and so on forever.

Cajori, Teaching and History of Mathematics in the U.S., Washington, 1890, 265–266. 89

A short, broad man of tremendous vitality, . . . Sylvester’s capacious head was ever lost in the highest cloud-lands of pure mathematics. Often in the dead of night he would get his favorite pupil, that he might communicate the very last product of his creative thought. Everything he saw suggested to him something new in the higher algebra. This transmutation of everything into new mathematics was a revelation to those who knew him intimately. They began to do it themselves.

Sylvester had one remarkable peculiarity. He seldom remembered theorems, propositions, etc., but had always to deduce them when he wished to use them. In this he was the very antithesis of Cayley, who was thoroughly conversant with everything that had been done in every branch of mathematics. I remember once submitting to Sylvester some investigations that I had been engaged on, and he immediately denied my first statement, saying that such a proposition had never been heard of, let alone proved. To his astonishment, I showed him a paper of his own in which he had proved the proposition; in fact, I believe the object of his paper had been the very proof which was so strange to him.

Another characteristic of Sylvester, which is very unusual among mathematicians, was his apparent inability to remember mathematics! W. P. Durfee had the following to say.

For more information about Sylvester, visit: http://www-groups.dcs.st-and .ac.uk/~history/

Sylvester’s enthusiasm for teaching and his influence on his students are captured in the following passage written by Sylvester’s first student at Johns Hopkins, G. B. Halsted.

90

91

4 | Supplementary Exercises for Chapters 1–4

Supplementary Exercises for Chapters 1–4 If you really want something in this life, you have to work for it—Now quiet, they’re about to announce the lottery numbers! HOMER SIMPSON

True/False questions for Chapters 1–4 are available on the web at: http://www.d.umn.edu/~jgallian/TF 1. Let G be a group and let H be a subgroup of G. For any fixed x in G, define xHx21 5 {xhx21 | h [ H}. Prove the following. a. xHx21 is a subgroup of G. b. If H is cyclic, then xHx21 is cyclic. c. If H is Abelian, then xHx21 is Abelian. The group xHx21 is called a conjugate of H. (Note that conjugation preserves structure.) 2. Let G be a group and let H be a subgroup of G. Define N(H) 5 {x [ G | xHx21 5 H}. Prove that N(H) (called the normalizer of H) is a subgroup of G.† 3. Let G be a group. For each a [ G, define cl(a) 5 {xax21 | x [ G}. Prove that these subsets of G partition G. [cl(a) is called the conjugacy class of a.] 4. The group defined by the following table is called the group of quaternions. Use the table to determine each of the following: a. The center b. cl(a) c. cl(b) d. All cyclic subgroups

e a a2 a3 b ba ba2 ba3

†This

e

a

a2

a3

b

ba

ba2

ba3

e a a2 a3 b ba ba2 ba3

a a2 a3 e ba ba2 ba3 b

a2 a3 e a ba2 ba3 b ba

a3 e a a2 ba3 b ba ba2

b ba3 ba2 ba a2 a e a3

ba b ba3 ba2 a3 a2 a e

ba2 ba b ba3 e a3 a2 a

ba3 ba2 ba b a e a3 a2

very important subgroup was first used by L. Sylow in 1872 to prove the existence of certain kinds of subgroups in a group. His work is discussed in Chapter 24.

92

Groups

5. (Conjugation preserves order.) Prove that, in any group, |xax21| 5 |a|. (This exercise is referred to in Chapter 24.) 6. Prove that, in any group, |ab| 5 |ba|. 7. If a, b, and c are elements of a group, give an example to show that it need not be the case that |abc| 5 |cba|. 8. Let a and b belong to a group G. Prove that there is an element x in G such that xax 5 b if and only if ab 5 c2 for some element c in G. 9. Prove that if a is the only element of order 2 in a group, then a lies in the center of the group. 10. Let G be the plane symmetry group of the infinite strip of equally spaced H’s shown below. H

H

H Axis 1

11. 12. 13. 14. 15.

16. 17.

18.

19. 20.

H

H

Axis 2

Let x be the reflection about Axis 1 and let y be the reflection about Axis 2. Calculate |x|, |y|, and |xy|. Must the product of elements of finite order have finite order? What are the orders of the elements of D15? How many elements have each of these orders? Prove that a group of order 4 is Abelian. Prove that a group of order 5 must be cyclic. Prove that an Abelian group of order 6 must be cyclic. Let G be an Abelian group and let n be a fixed positive integer. Let Gn 5 {gn | g [ G}. Prove that Gn is a subgroup of G. Give an example showing that Gn need not be a subgroup of G when G is non-Abelian. (This exercise is referred to in Chapter 11.) Let G 5 5a 1 b"2 6 , where a and b are rational numbers not both 0. Prove that G is a group under ordinary multiplication. (1969 Putnam Competition) Prove that no group is the union of two proper subgroups. Does the statement remain true if “two” is replaced by “three”? Prove that the subset of elements of finite order in an Abelian group forms a subgroup. (This subgroup is called the torsion subgroup.) Is the same thing true for non-Abelian groups? Let p be a prime and let G be an Abelian group. Show that the set of all elements whose orders are powers of p is a subgroup of G. Suppose that a and b are group elements. If |b| 5 2 and bab 5 a4 , determine the possibilities for |a| .

4 | Supplementary Exercises for Chapters 1–4

93

21. Suppose that a finite group is generated by two elements a and b (that is, every element of the group can be expressed as some product of a’s and b’s). Given that a35 b2 5 e and ba2 5 ab, construct the Cayley table for the group. We have already seen an example of a group that satisfies these conditions. Name it. 22. If a is an element of a group and |a| 5 n, prove that C(a) 5 C(ak ) when k is relatively prime to n. 23. Let x and y belong to a group G. If xy [ Z(G), prove that xy 5 yx. 24. Suppose that H and K are nontrivial subgroups of Q under addition. Show that H > K is a nontrivial subgroup of Q. Is this true if Q is replaced by R? 25. Let H be a subgroup of G and let g be an element of G. Prove that N(gHg21) 5 gN(H)g21. See Exercise 2 for the notation. 26. Let H be a subgroup of a group G and let |g| 5 n. If gm belongs to H and m and n are relatively prime, prove that g belongs to H. 27. Find a group that contains elements a and b such that |a| 5 2, |b| 5 11, and |ab| 5 2. 28. Suppose that G is a group with exactly eight elements of order 10. How many cyclic subgroups of order 10 does G have? 29. (1989 Putnam Competition) Let S be a nonempty set with an associative operation that is left and right cancellative (xy 5 xz implies y 5 z, and yx 5 zx implies y 5 z). Assume that for every a in S the set {an | n 5 1, 2, 3, . . .} is finite. Must S be a group? 30. Let H1, H2, H3, . . . be a sequence of subgroups of a group with the property that H1 # H2 # H3 . . . . Prove that the union of the sequence is a subgroup. 31. Let R* be the group of nonzero real numbers under multiplication and let H 5{g [ R*| some nonzero integer power of g is a rational number}. Prove that H is a subgroup of R*. 32. Suppose that a and b belong to a group, a and b commute, and |a| and |b| are relatively prime. Prove that |ab| 5 |a||b|. Give an example showing that |ab| need not be |a||b| when a and b commute but |a| and |b| are not relatively prime. (Don’t use a [ b.) 33. Let H 5 {A [ GL(2, R) | det A is rational}. Prove or disprove that H is a subgroup of GL(2, R). What if “rational” is replaced by “an integer”? 34. Suppose that G is a group that has exactly one nontrivial proper subgroup. Prove that G is cyclic and |G| 5 p2, where p is prime. 35. Suppose that G is a group and G has exactly two nontrivial proper subgroups. Prove that G is cyclic and |G| 5 pq, where p and q are distinct primes, or that G is cyclic and |G| 5 p3, where p is prime.

94

Groups

36. If |a2| 5 |b2|, prove or disprove that |a| 5 |b|. 37. (1995 Putnam Competition) Let S be a set of real numbers that is closed under multiplication. Let T and U be disjoint subsets of S whose union is S. Given that the product of any three (not necessarily distinct) elements of T is in T and that the product of any three elements of U is in U, show that at least one of the two subsets T and U is closed under multiplication. 38. If p is an odd prime, prove that there is no group that has exactly p elements of order p. 39. Give an example of a group G with infinitely many distinct subgroups H1, H2, H3, . . . such that H1 , H2 , H3 . . . . 40. Suppose a and b are group elements and b 2 e. If a21ba 5 b2 and |a| 5 3, find |b|. What is |b|, if |a| 5 5? What can you say about |b| in the case where |a| 5 k? 41. Let a and b belong to a group G. Show that there is an element g in G such that g21 abg 5 ba. 42. Suppose G is a group and x3y3 5 y3x3 for every x and y in G. Let H 5 {x [ G| |x| is relatively prime to 3}. Prove that elements of H commute with each other and that H is a subgroup of G. Is your argument valid if 3 is replaced by an arbitrary positive integer n? Explain why or why not. 43. Let G be a finite group and let S be a subset of G that contains more than half of the elements of G. Show that every element of G can be expressed in the form s1s2 where s1 and s2 belong to S. 44. Let G be a group and let f be a function from G to some set. Show that H 5 {g [ G| f (xg) 5 f (x) for all x [ G} is a subgroup of G. In the case that G is the group of real numbers under addition and f (x) 5 sin x, describe H. 45. Let G be a cyclic group of order n and let H be the subgroup of order d. Show that H 5 {x [ G| |x| divides d}. 46. Let a be an element of maximum order from a finite Abelian group G. Prove that for any element b in G, |b| divides |a|. Show by example that this need not be true for finite non-Abelian groups. 47. Define an operation * on the set of integers by a * b 5 a 1 b 2 1. Show that the set of integers under this operation is a cyclic group. 48. Let n be an integer greater than 1. Find a noncyclic subgroup of U(4n) of order 4 that contains the element 2n 2 1.

5 Permutation Groups

Wigner’s discovery about the electron permutation group was just the beginning. He and others found many similar applications and nowadays group theoretical methods—especially those involving characters and representations—pervade all branches of quantum mechanics. GEORGE MACKEY, Proceedings of the

American Philosophical Society

Definition and Notation In this chapter, we study certain groups of functions, called permutation groups, from a set A to itself. In the early and mid-19th century, groups of permutations were the only groups investigated by mathematicians. It was not until around 1850 that the notion of an abstract group was introduced by Cayley, and it took another quarter century before the idea firmly took hold. Definitions Permutation of A, Permutation Group of A

A permutation of a set A is a function from A to A that is both oneto-one and onto. A permutation group of a set A is a set of permutations of A that forms a group under function composition.

Although groups of permutations of any nonempty set A of objects exist, we will focus on the case where A is finite. Furthermore, it is customary, as well as convenient, to take A to be a set of the form {1, 2, 3, . . . , n} for some positive integer n. Unlike in calculus, where most functions are defined on infinite sets and are given by formulas, in algebra, permutations of finite sets are usually given by an explicit listing of each element of the domain and its corresponding functional value. For example, we define a permutation a of the set {1, 2, 3, 4} by specifying a(1) 5 2,

a(2) 5 3,

a(3) 5 1,

a(4) 5 4.

95

96

Groups

A more convenient way to express this correspondence is to write a in array form as a 5 c

1 2 3 4 d. 2 3 1 4

Here a( j) is placed directly below j for each j. Similarly, the permutation b of the set {1, 2, 3, 4, 5, 6} given by b(1) 5 5, b(2) 5 3,

b(3) 5 1,

b(4) 5 6,

b(5) 5 2,

b(6) 5 4

is expressed in array form as b5c

1 2 3 4 5 6 d. 5 3 1 6 2 4

Composition of permutations expressed in array notation is carried out from right to left by going from top to bottom, then again from top to bottom. For example, let s5c

1 2 3 4 5 d 2 4 3 5 1

and g 5 c

1 2 3 4 5 d; 5 4 1 2 3

then gs 5 £

1 2 3 4 5 5 4 1 2 3

§ £

1 2 3 4 5 2 4 3 5 1

§ 5 c

1 2 3 4 5 d 4 2 1 3 5

On the right we have 4 under 1, since (gs)(1) 5 g(s(1)) 5 g(2) 5 4, so gs sends 1 to 4. The remainder of the bottom row gs is obtained in a similar fashion. We are now ready to give some examples of permutation groups. EXAMPLE 1 Symmetric Group S3 Let S3 denote the set of all oneto-one functions from {1, 2, 3} to itself. Then S3, under function composition, is a group with six elements. The six elements are 1 2 3 e5c d, 1 2 3

a5c

1 2 3 d, 2 3 1

a2 5 c

1 2 3 d, 3 1 2

5 | Permutation Groups

b5 c

1 2 3 d, 1 3 2

Note that ba 5 c

ab 5 c

1 2 3 d, 2 1 3

a2b 5 c

97

1 2 3 d. 3 2 1

1 2 3 d 5 a 2b 2 ab, so that S3 is non-Abelian. 3 2 1

The relation ba 5 a 2b can be used to compute other products in S3 without resorting to the arrays. For example, ba2 5 (ba)a 5 (a 2b)a 5 a 2(ba) 5 a 2(a 2b) 5 a 4b 5 ab. Example 1 can be generalized as follows. EXAMPLE 2 Symmetric Group Sn Let A 5 {1, 2, . . . , n}. The set of all permutations of A is called the symmetric group of degree n and is denoted by Sn. Elements of Sn have the form a5 c

1 2 c n d. a(1) a(2) ca(n)

It is easy to compute the order of Sn. There are n choices of a(1). Once a(1) has been determined, there are n 2 1 possibilities for a(2) [since a is one-to-one, we must have a(1) 2 a(2)]. After choosing a(2), there are exactly n 2 2 possibilities for a(3). Continuing along in this fashion, we see that Sn has n(n 2 1) ? ? ? 3 ? 2 ? 1 5 n! elements. We leave it to the reader to prove that Sn is non-Abelian when n $ 3 (Exercise 41). The symmetric groups are rich in subgroups. The group S4 has 30 subgroups, and S5 has well over 100 subgroups. EXAMPLE 3 Symmetries of a Square As a third example, we associate each motion in D4 with the permutation of the locations of each of the four corners of a square. For example, if we label the four corner positions as in the figure below and keep these labels fixed for reference, we may describe a 90° counterclockwise rotation by the permutation 3

2

4

1

r 5 c

1 2 3 4 d, 2 3 4 1

98

Groups

whereas a reflection across a horizontal axis yields f 5 c

1 2 3 4 d. 2 1 4 3

These two elements generate the entire group (that is, every element is some combination of the r’s and f’s). When D4 is represented in this way, we see that it is a subgroup of S4.

Cycle Notation There is another notation commonly used to specify permutations. It is called cycle notation and was first introduced by the great French mathematician Cauchy in 1815. Cycle notation has theoretical advantages in that certain important properties of the permutation can be readily determined when cycle notation is used. As an illustration of cycle notation, let us consider the permutation a5c

1 2 3 4 5 6 d. 2 1 4 6 5 3

This assignment of values could be presented schematically as follows: 1

3

5

α α

α

α 6 2

4

α

α

Although mathematically satisfactory, such diagrams are cumbersome. Instead, we leave out the arrows and simply write a 5 (1, 2) (3, 4, 6)(5). As a second example, consider 1 2 3 4 5 6 b5c d. 5 3 1 6 2 4 In cycle notation, b can be written (2, 3, 1, 5)(6, 4) or (4, 6)(3, 1, 5, 2), since both of these unambiguously specify the function b. An expression of the form (a1, a2, . . . , am) is called a cycle of length m or an m-cycle.

5 | Permutation Groups

99

A multiplication of cycles can be introduced by thinking of a cycle as a permutation that fixes any symbol not appearing in the cycle. Thus, the cycle (4, 6) can be thought of as representing the 1 2 3 4 5 6 d . In this way, we can multiply cycles permutation c 1 2 3 6 5 4 by thinking of them as permutations given in array form. Consider the following example from S8. Let a 5 (13)(27)(456)(8) and b 5 (1237)(648)(5). (When the domain consists of single-digit integers, it is common practice to omit the commas between the digits.) What is the cycle form of ab? Of course, one could say that ab 5 (13)(27)(456)(8)(1237)(648)(5), but it is usually more desirable to express a permutation in a disjoint cycle form (that is, the various cycles have no number in common). Well, keeping in mind that function composition is done from right to left and that each cycle that does not contain a symbol fixes the symbol, we observe that: (5) fixes 1; (648) fixes 1; (1237) sends 1 to 2; (8) fixes 2; (456) fixes 2; (27) sends 2 to 7; and (13) fixes 7. So the net effect of ab is to send 1 to 7. Thus we begin ab 5 (17 ? ? ?) ? ? ? . Now, repeating the entire process beginning with 7, we have, cycle by cycle, right to left, 7 → 7 → 7 → 1 → 1 → 1 → 1 → 3, so that ab 5 (173 ? ? ?) ? ? ? . Ultimately, we have ab 5 (1732)(48)(56). The important thing to bear in mind when multiplying cycles is to “keep moving” from one cycle to the next from right to left. (Warning: Some authors compose cycles from left to right. When reading another text, be sure to determine which convention is being used.) To be sure you understand how to switch from one notation to the other and how to multiply permutations, we will do one more example of each. If array notations for a and b, respectively, are c

1 2 3 4 5 d 2 1 3 5 4

and

c

1 2 3 4 5 d 5 4 1 2 3

then, in cycle notation, a 5 (12)(3)(45), b 5 (153)(24), and ab 5 (12)(3)(45)(153)(24). To put ab in disjoint cycle form, observe that (24) fixes 1; (153) sends 1 to 5; (45) sends 5 to 4; and (3) and (12) both fix 4. So, ab sends 1 to 4. Continuing in this way we obtain ab 5 (14)(253). One can convert ab back to array form without converting each cycle of ab into array form by simply observing that (14) means 1 goes to 4 and 4 goes to 1; (253) means 2 → 5, 5 → 3, 3 → 2. One final remark about cycle notation: Mathematicians prefer not to write cycles that have only one entry. In this case, it is understood that any

100

Groups

missing element is mapped to itself. With this convention, the permutation a above can be written as (12)(45). Similarly, a5c

1 2 3 4 5 d 3 2 4 1 5

can be written a 5 (134). Of course, the identity permutation consists only of cycles with one entry, so we cannot omit all of these! In this case, one usually writes just one cycle. For example, e5c

1 2 3 4 5 d 1 2 3 4 5

can be written as e 5 (5) or e 5 (1). Just remember that missing elements are mapped to themselves.

Properties of Permutations We are now ready to state several theorems about permutations and cycles. The proof of the first theorem is implicit in our discussion of writing permutations in cycle form. Theorem 5.1 Products of Disjoint Cycles Every permutation of a finite set can be written as a cycle or as a product of disjoint cycles.

PROOF Let a be a permutation on A 5 {1, 2, . . . , n}. To write a in disjoint cycle form, we start by choosing any member of A, say a1, and let a2 5 a(a1),

a3 5 a(a(a1)) 5 a 2(a1),

and so on, until we arrive at a1 5 a m(a1) for some m. We know that such an m exists because the sequence a1, a(a1), a 2(a1), ? ? ? must be finite; so there must eventually be a repetition, say a i(a1) 5 a j(a1) for some i and j with i , j. Then a1 5 a m(a1), where m 5 j 2 i. We express this relationship among a1, a2, . . . , am as a 5 (a1, a2, . . . , am) ? ? ? . The three dots at the end indicate the possibility that we may not have exhausted the set A in this process. In such a case, we merely choose any element b1 of A not appearing in the first cycle and proceed to create a new cycle as before. That is, we let b2 5 a(b1), b3 5 a2(b1), and so on, until we reach b1 5 a k(b1) for some k. This new cycle will have no

5 | Permutation Groups

101

elements in common with the previously constructed cycle. For, if so, then a i(a1) 5 a j(b1) for some i and j. But then a i2j(a1) 5 b1, and therefore b1 5 at for some t. This contradicts the way b1 was chosen. Continuing this process until we run out of elements of A, our permutation will appear as a 5 (a1, a2, . . . , am)(b1, b2, . . . , bk) ? ? ? (c1, c2, . . . , cs). In this way, we see that every permutation can be written as a product of disjoint cycles. Theorem 5.2 Disjoint Cycles Commute If the pair of cycles a 5 (a1, a2, . . . , am) and b 5 (b1,b2, . . . , bn ) have no entries in common, then ab 5 ba.

PROOF For definiteness, let us say that a and b are permutations of the set S 5 {a1, a2, . . . , am, b1, b2, . . . , bn, c1, c2, . . . , ck} where the c’s are the members of S left fixed by both a and b (there may not be any c’s). To prove that ab 5 ba, we must show that (ab)(x) 5 (ba)(x) for all x in S. If x is one of the a elements, say ai, then (ab)(ai) 5 a(b(ai)) 5 a(ai) 5 ai11, since b fixes all a elements. (We interpret ai11 as a1 if i 5 m.) For the same reason, (ba)(ai) 5 b(a(ai)) 5 b(ai11) 5 ai11. Hence, the functions of ab and ba agree on the a elements. A similar argument shows that ab and ba agree on the b elements as well. Finally, suppose that x is a c element, say ci. Then, since both a and b fix c elements, we have (ab)(ci) 5 a(b(ci)) 5 a(ci) 5 ci and (ba)(ci) 5 b(a(ci)) 5 b(ci) 5 ci . This completes the proof. In demonstrating how to multiply cycles, we showed that the product (13)(27)(456)(8)(1237)(648)(5) can be written in disjoint cycle

102

Groups

form as (1732)(48)(56). Is economy in expression the only advantage to writing a permutation in disjoint cycle form? No. The next theorem shows that the disjoint cycle form has the enormous advantage of allowing us to “eyeball” the order of the permutation. Theorem 5.3 Order of a Permutation (Ruffini—1799) The order of a permutation of a finite set written in disjoint cycle form is the least common multiple of the lengths of the cycles.

PROOF First, observe that a cycle of length n has order n. (Verify this yourself.) Next, suppose that a and b are disjoint cycles of lengths m and n, and let k be the least common multiple of m and n. It follows from Theorem 4.1 that both ak and bk are the identity permutation e and, since a and b commute, (ab)k 5 a kb k is also the identity. Thus, we know by Corollary 2 to Theorem 4.1 (ak 5 e implies that |a| divides k) that the order of ab—let us call it t—must divide k. But then (ab)t 5 a tb t 5 e, so that a t 5 b 2t. However, it is clear that if a and b have no common symbol, the same is true for a t and b 2t, since raising a cycle to a power does not introduce new symbols. But, if a t and b 2t are equal and have no common symbol, they must both be the identity, because every symbol in a t is fixed by b 2t and vice versa (remember that a symbol not appearing in a permutation is fixed by the permutation). It follows, then, that both m and n must divide t. This means that k, the least common multiple of m and n, divides t also. This shows that k 5 t. Thus far, we have proved that the theorem is true in the cases where the permutation is a single cycle or a product of two disjoint cycles. The general case involving more than two cycles can be handled in an analogous way. Theorem 5.3 is an enomously powerful tool for calculating the orders of permuations. We demonstrate this in the next example. EXAMPLE 4 To determine the orders of the 5040 elements of S7 , we need only consider the possible disjoint cycle structures of the elements of S7 . For convenience, we denote an n-cycle by (n). Then, arranging all possible disjoint cycle structures of elements of S7 according to longest cycle lengths left to right, we have

5 | Permutation Groups

103

(7) (6) (1) (5) (2) (5) (1) (1) (4) (3) (4) (2) (1) (4) (1) (1) (1) (3) (3) (1) (3) (2) (2) (3) (2) (1) (1) (3) (1) (1) (1) (1) (1) (2) (2) (2) (1) (2) (2) (1) (1) (1) (2) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1). Now, from Theorem 5.3 we see that the orders of the elements of S7 are 7, 6, 10, 5, 12, 4, 3, 2, and 1. To do the same for the 10! 5 3,628,800 elements of S10 would be nearly as simple. As we will soon see, a particularly important kind of permutation is a cycle of length 2—that is, a permutation of the form (ab) where a 2 b. Many authors call these permutations transpositions, since the effect of (ab) is to interchange or transpose a and b. Theorem 5.4 Product of 2-Cycles Every permutation in Sn, n . 1, is a product of 2-cycles.

PROOF First, note that the identity can be expressed as (12)(12), and so it is a product of 2-cycles. By Theorem 5.1, we know that every permutation can be written in the form (a1a2 ? ? ? ak)(b1b2 ? ? ? bt) ? ? ? (c1c2 ? ? ? cs). A direct computation shows that this is the same as (a1ak)(a1ak21) ? ? ? (a1a2)(b1bt)(b1bt21) ? ? ? (b1b2) ? ? ? (c1cs)(c1cs21) ? ? ? (c1c2). This completes the proof. The decompositions in the following example demonstrate this technique.

104

Groups

EXAMPLE 5 (12345) 5 (15)(14)(13)(12) (1632)(457) 5 (12)(13)(16)(47)(45) The decomposition of a permutation into a product of 2-cycles given in the proof of Theorem 5.4 is not the only way a permutation can be written as a product of 2-cycles. Although the next example shows that even the number of 2-cycles may vary from one decomposition to another, we will prove in Theorem 5.5 (first proved by Cauchy) that there is one aspect of a decomposition that never varies. EXAMPLE 6 (12345) 5 (54)(53)(52)(51) (12345) 5 (54)(52)(21)(25)(23)(13) We isolate a special case of Theorem 5.5 as a lemma. Lemma If e 5 b1b2 ? ? ? br , where the b’s are 2-cycles, then r is even.

PROOF Clearly, r 2 1, since a 2-cycle is not the identity. If r 5 2, we are done. So, we suppose that r . 2, and we proceed by induction. Since (ij) 5 ( ji), the product br21br can be expressed in one of the following forms shown on the right: e 5 (ab)(ab) (ab)(bc) 5 (ac)(ab) (ac)(cb ) 5 (bc)(ab) (ab)(cd) 5 (cd)(ab). If the first case occurs, we may delete br21 br from the original product to obtain e 5 b1 b2 ? ? ? br22. In the other three cases, we replace the form of br21 br on the right by its counterpart on the left to obtain a new product of r 2-cycles that is still the identity, but where the rightmost occurrence of the integer a is in the second-from-the-rightmost 2-cycle of the product instead of the rightmost 2-cycle. We now repeat the procedure just described with br22 br21, and, as before, we obtain a product of (r 2 2) 2-cycles equal to the identity or a new product of r 2-cycles, where the rightmost occurrence of a is in the third 2-cycle from the right.

5 | Permutation Groups

105

Continuing this process, we must obtain a product of (r 2 2) 2-cycles equal to the identity, because otherwise we have a product equal to the identity in which the only occurrence of the integer a is in the leftmost 2cycle, and such a product does not fix a, whereas the identity does. Hence, by the Second Principle of Mathematical Induction, r 2 2 is even, and r is even as well. Theorem 5.5 Always Even or Always Odd If a permutation a can be expressed as a product of an even (odd) number of 2-cycles, then every decomposition of a into a product of 2-cycles must have an even (odd) number of 2-cycles. In symbols, if

a 5 b1b2 ? ? ? br

and

a 5 g1g2 ? ? ? gs ,

where the b’s and the g’s are 2-cycles, then r and s are both even or both odd.

PROOF Observe that b1b2 ? ? ? br 5 g1g2 ? ? ? gs implies e 5 g1g2 ? ? ? gsbr 21 ? ? ? b221b121 5 g1g2 ? ? ? gsbr ? ? ? b2b1, since a 2-cycle is its own inverse. Thus, the lemma on page 104 guarantees that s 1 r is even. It follows that r and s are both even or both odd. Definition Even and Odd Permutations

A permutation that can be expressed as a product of an even number of 2-cycles is called an even permutation. A permutation that can be expressed as a product of an odd number of 2-cycles is called an odd permutation.

Theorems 5.4 and 5.5 together show that every permutation can be unambiguously classified as either even or odd. The significance of this observation is given in Theorem 5.6. Theorem 5.6 Even Permutations Form a Group The set of even permutations in Sn forms a subgroup of Sn.

PROOF This proof is left to the reader (Exercise 13). The subgroup of even permutations in Sn arises so often that we give it a special name and notation.

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Groups

Definition Alternating Group of Degree n

The group of even permutations of n symbols is denoted by An and is called the alternating group of degree n.

The next result shows that exactly half of the elements of Sn(n . 1) are even permutations. Theorem 5.7 For n . 1, An has order n!/2.

PROOF For each odd permutation a, the permutation (12)a is even and (12)a 2 (12)b when a 2 b. Thus, there are at least as many even permutations as there are odd ones. On the other hand, for each even permutation a, the permutation (12)a is odd and (12)a 2 (12)b when a 2 b. Thus, there are at least as many odd permutations as there are even ones. It follows that there are equal numbers of even and odd permutations. Since |Sn| 5 n!, we have |An| 5 n!/2. The names for the symmetric group and the alternating group of degree n come from the study of polynomials over n variables. A symmetric polynomial in the variables x1, x2, . . . , xn is one that is unchanged under any transposition of two of the variables. An alternating polynomial is one that changes signs under any transposition of two of the variables. For example, the polynomial x1 x2 x3 is unchanged by any transposition of two of the three variables, whereas the polynomial (x12x2)(x12x3)(x22x3) changes signs when any two of the variables are transposed. Since every member of the symmetric group is the product of transpositions, the symmetric polynomials are those that are unchanged by members of the symmetric group. Likewise, since any member of the alternating group is the product of an even number of transpositions, the alternating polynomials are those that are unchanged by members of the alternating group and change sign by the other permutations of Sn. The alternating groups are among the most important examples of groups. The groups A4 and A5 will arise on several occasions in later chapters. In particular, A5 has great historical significance. A geometric interpretation of A4 is given in Example 7, and a multiplication table for A4 is given as Table 5.1. EXAMPLE 7 ROTATIONS OF A TETRAHEDRON The 12 rotations of a regular tetrahedron can be conveniently described with the elements of A4. The top row of Figure 5.1 illustrates the identity and three 180° “edge” rotations about axes joining midpoints of two edges.

107

5 | Permutation Groups

Table 5.1 The Alternating Group A4 of Even Permutations of {1, 2, 3, 4} (In this table, the permutations of A4 are designated as a1, a2, . . . , a12 and an entry k inside the table represents ak. For example, a3 a8 5 a6.) (1) 5 a1 (12)(34) 5 a2 (13)(24) 5 a3 (14)(23) 5 a4 (123) 5 a5 (243) 5 a6 (142) 5 a7 (134) 5 a8 (132) 5 a9 (143) 5 a10 (234) 5 a11 (124) 5 a12

a1

a2

a3

a4

a5

a6

a7

a8

a9

a10

a11

a12

1 2 3 4 5 6 7 8 9 10 11 12

2 1 4 3 8 7 6 5 11 12 9 10

3 4 1 2 6 5 8 7 12 11 10 9

4 3 2 1 7 8 5 6 10 9 12 11

5 6 7 8 9 10 11 12 1 2 3 4

6 5 8 7 12 11 10 9 3 4 1 2

7 8 5 6 10 9 12 11 4 3 2 1

8 7 6 5 11 12 9 10 2 1 4 3

9 10 11 12 1 2 3 4 5 6 7 8

10 9 12 11 4 3 2 1 7 8 5 6

11 12 9 10 2 1 4 3 8 7 6 5

12 11 10 9 3 4 1 2 6 5 8 7

1

(1)

B

A A

2

1

(12)(34)

D 4

B

C 4

C C

1 C

3

3

1 D

C 4 D

1 (234)

D AA 3

3

(142)

2

B 4

C D

2

C C

3

3

B B 3

2

C B 4

A C C

1

(143)

D D C 4

2 D

A 4

D

D

1 A

1

B B

(124)

4

2 C

4

1 A

B B

BB

1

3

3

A 4

2 C

A A

D

A A

BB

(132)

(243)

C

B

2

B 4

D A A

(134)

2 A

DD

DD

3

(123)

2

1

(14)(23) C C

B

A

2

1

(13)(24)

2

3

Figure 5.1 Rotations of a regular tetrahedron.

B

C A

D D 3

4

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Groups

The second row consists of 120° “face” rotations about axes joining a vertex to the center of the opposite face. The third row consists of 2120° (or 240°) “face” rotations. Notice that the four rotations in the second row can be obtained from those in the first row by left-multiplying the four in the first row by the rotation (123), whereas those in the third row can be obtained from those in the first row by left-multiplying the ones in the first row by (132). Many molecules with chemical formulas of the form AB4, such as methane (CH4) and carbon tetrachloride (CCl4), have A4 as their symmetry group. Figure 5.2 shows the form of one such molecule. Many games and puzzles can be analyzed using permutations.

Figure 5.2 A tetrahedral AB4 molecule.

EXAMPLE 8 (Loren Larson) A Sliding Disk Puzzle Consider the puzzle shown below (the space in the middle is empty). 1 6

2

5

3 4

By sliding disks from one position to another along the lines indicated without lifting or jumping, can we obtain the following arrangement?

5 | Permutation Groups

109

3 1

5

6

2 4

To answer this question, we view the positions as numbered in the first figure above and consider two basic operations. Let r denote the following operation: Move the disk in position 1 to the center position, then move the disk in position 6 to position 1, the disk in position 5 to position 6, the disk in position 4 to position 5, the disk in position 3 to position 4, then the disk in the middle position to position 3. Let s denote the operation: Move the disk in position 1 to the center position, then move the disk in position 2 to position 1, then move the disk in position 3 to position 2, and finally move the disk in the center to position 3. In permutation notation, we have r 5 (13456) and s 5 (132). The permutation for the arrangement we seek is (16523). Clearly, if we can express (16523) as a string of r’s and s’s, we can achieve the desired arangement. Rather than attempt to find an appropriate combination of r’s and s’s by hand, it is easier to employ computer software that is designed for this kind of problem. One such software program is GAP (see Suggested Software at the end of this chapter). With GAP, all we need to do is use the following commands: gap. G :5 SymmetricGroup(6); gap. r :5 (1,3,4,5,6); s :5 (1, 3, 2); gap. K :5 Subgroup(G,[r,s]); gap. Factorization(K,(1,6,5,2,3)); The first three lines inform the computer that our group is the subgroup of S6 generated by r 5 (13456) and s 5 (132). The fourth line requests that (16523) be expressed in terms of r and s. The response to the command gap. Size (K); tells us that the order of the subgroup generated by r and s is 360. Then, observing that r and s are even permutations and that |A6| 5 360, we deduce that r and s can achieve any arrangement that corresponds to an even permutation.

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Groups

GAP can even compute the 43,252,003,274,489,856,000 (431 quintillion) permutations of the Rubik’s Cube! Labeling the faces of the cube as shown here,

9 12 14

10 left 15

11 13 16

1 4 6

2 top 7

3 5 8

17 20 22

18 front 23

19 21 24

42 bottom 47

43 45 48

41 44 46

25 28 30

26 right 31

27 29 32

33 36 38

34 rear 39

35 37 40

the group of permutations of the cube is generated by the following rotations of the six layers: top 5 (1,3,8,6)(2,5,7,4)(9,33,25,17)(10,34,26,18)(11,35,27,19) left 5 (9,11,16,14)(10,13,15,12)(1,17,41,40)(4,20,44,37)(6,22,46,35) front 5 (17,19,24,22)(18,21,23,20)(6,25,43,16)(7,28,42,13)(8,30,41,11) right 5 (25,27,32,30)(26,29,31,28)(3,38,43,19)(5,36,45,21)(8,33,48,24) rear 5 (33,35,40,38)(34,37,39,36)(3,9,46,32)(2,12,47,29)(1,14,48,27) bottom 5 (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39) (16,24,32,40)

A Check Digit Scheme Based on D5 In Chapter 0, we presented several schemes for appending a check digit to an identification number. Among these schemes, only the International Standard Book Number method was capable of detecting all single-digit errors and all transposition errors involving adjacent digits. However, recall that this success was achieved by introducing the alphabetical character X to handle the case where 10 was required to make the dot product 0 modulo 11. In contrast, in 1969, J. Verhoeff [2] devised a method utilizing the dihedral group of order 10 that detects all single-digit errors and all transposition errors involving adjacent digits without the necessity of avoiding certain numbers or introducing a new character. To describe this method, consider the permutation s 5 (01589427)(36) and the dihedral group of order 10 as represented in Table 5.2. (Here we use 0 through 4 for the rotations, 5 through 9 for the reflections, and * for the operation of D5.)

5 | Permutation Groups

111

Table 5.2 Multiplication for D5 *

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9

1 2 3 4 0 9 5 6 7 8

2 3 4 0 1 8 9 5 6 7

3 4 0 1 2 7 8 9 5 6

4 0 1 2 3 6 7 8 9 5

5 6 7 8 9 0 1 2 3 4

6 7 8 9 5 4 0 1 2 3

7 8 9 5 6 3 4 0 1 2

8 9 5 6 7 2 3 4 0 1

9 5 6 7 8 1 2 3 4 0

Verhoeff’s idea is to view the digits 0 through 9 as the elements of the group D5 and to replace ordinary addition with calculations done in D5. In particular, to any string of digits a1a2 . . . an21, we append the check digit an so that s(a1) p s 2 (a2 ) p ? ? ? p s n 22(an22) p s n 21(an21) p s n (an ) 5 0. [Here s 2 (x) 5 s(s(x)), s 3 (x) 5 s(s 2 (x)), and so on.] Since s has the property that s i (a) 2 s i (b) if a 2 b, all single-digit errors are detected. Also, because a p s(b) 2 b p s(a)

if a 2 b,

(1)

as can be checked on a case-by-case basis (see Exercise 49), it follows that all transposition errors involving adjacent digits are detected [since Equation (1) implies that s i(a) p s i11(b) 2 si(b) p s i11(a) if a 2 b]. From 1990 until 2002, the German government used a minor modification of Verhoeff’s check-digit scheme to append a check digit to the serial numbers on German banknotes. Table 5.3 gives the values of the functions s, s 2 , . . . , s10 needed for the computations. [The functional value s i ( j) appears in the row labeled with s i and the column labeled j.] Since the serial numbers on the banknotes use 10 letters of the alphabet in addition to the 10 decimal digits, it is necessary to assign numerical values to the letters to compute the check digit. This assignment is shown in Table 5.4. To any string of digits a1a2 . . . a10 corresponding to a banknote serial number, the check digit a11 is chosen such that s (a1) p s 2(a2) p ? ? ? p s9(a9) p s10(a10) p a11 5 0 [instead of s(a1) p s 2 (a2) p ? ? ? p s10(a10) p s11(a11) 5 0 as in the Verhoeff scheme].

112

Groups

Table 5.3 Powers of s

s s2 s3 s4 s5 s6 s7 s8 s9 s10

0

1

2

3

4

5

6

7

8

9

1 5 8 9 4 2 7 0 1 5

5 8 9 4 2 7 0 1 5 8

7 0 1 5 8 9 4 2 7 0

6 3 6 3 6 3 6 3 6 3

2 7 0 1 5 8 9 4 2 7

8 9 4 2 7 0 1 5 8 9

3 6 3 6 3 6 3 6 3 6

0 1 5 8 9 4 2 7 0 1

9 4 2 7 0 1 5 8 9 4

4 2 7 0 1 5 8 9 4 2

Table 5.4 Letter Values A

D

G

K

L

N

S

U

Y

Z

0

1

2

3

4

5

6

7

8

9

To trace through a specific example, consider the banknote (featuring the mathematician Gauss) shown in Figure 5.3 with the number AG8536827U7. To verify that 7 is the appropriate check digit, we observe that s(0) p s 2 (2) p s 3 (8) p s 4 (5) p s 5 (3) p s 6 (6) p s 7 (8) p s 8 (2) p s 9 (7) p s 10(7) p 7 5 1 p 0 p 2 p 2 p 6 p 6 p 5 p 2 p 0 p 1 p 7 5 0, as it should be. [To illustrate how to use the multiplication table for D5, we compute 1 p 0 p 2 p 2 5 (1 p 0) p 2 p 2 5 1 p 2 p 2 5 (1 p 2) p 2 5 3 p 2 5 0.]

Figure 5.3 German banknote with serial number AG8536827U and check digit 7.

One shortcoming of the German banknote scheme is that it does not distinguish between a letter and its assigned numerical value. Thus, a

5 | Permutation Groups

113

substitution of 7 for U (or vice versa) and the transposition of 7 and U are not detected by the check digit. Moreover, the banknote scheme does not detect all transpositions of adjacent characters involving the check digit itself. For example, the transposition of D and 8 in positions 10 and 11 is not detected. Both of these defects can be avoided by using the Verhoeff method with D18, the dihedral group of order 36, to assign every letter and digit a distinct value together with an appropriate function s (see Gallian [1]). Using this method to append a check character, all single-position errors and all transposition errors involving adjacent digits will be detected.

Exercises

Text not available due to copyright restrictions

1. Find the order of each of the following permutations. a. (14) b. (147) c. (14762) d. (a1a2 . . . ak ) 2. Write each of the following permutations as a product of disjoint cycles. a. (1235)(413) b. (13256)(23)(46512) c. (12)(13)(23)(142) 3. What is the order of each of the following permutations? a. (124)(357) b. (124)(3567) c. (124)(35) d. (124)(357869) e. (1235)(24567) f. (345)(245)

114

Groups

4. What is the order of each of the following permutations? 1 2 3 4 5 6 d a. c 2 1 5 4 6 3 1 2 3 4 5 6 7 d 7 6 1 2 3 4 5 What is the order of the product of a pair of disjoint cycles of lengths 4 and 6? Show that A8 contains an element of order 15. What are the possible orders for the elements of S6 and A6? What about A7? (This exercise is referred to in Chapter 25.) What is the maximum order of any element in A10? Determine whether the following permutations are even or odd. a. (135) b. (1356) c. (13567) d. (12)(134)(152) e. (1243)(3521) Show that a function from a finite set S to itself is one-to-one if and only if it is onto. Is this true when S is infinite? (This exercise is referred to in Chapter 6.) Let n be a positive integer. If n is odd, is an n-cycle an odd or an even permutation? If n is even, is an n-cycle an odd or an even permutation? If a is even, prove that a21 is even. If a is odd, prove that a21 is odd. Prove Theorem 5.6. In Sn, let a be an r-cycle, b an s-cycle, and g a t-cycle. Complete the following statements: ab is even if and only if r 1 s is ______; abg is even if and only if r 1 s 1 t is ______. Let a and b belong to Sn. Prove that ab is even if and only if a and b are both even or both odd. Associate an even permutation with the number 11 and an odd permutation with the number 21. Draw an analogy between the result of multiplying two permutations and the result of multiplying their corresponding numbers 11 or 21. Let 1 2 3 4 5 6 1 2 3 4 5 6 a5 c d and b 5 c d. 2 1 3 5 4 6 6 1 2 4 3 5

b. c 5. 6. 7. 8. 9.

10.

11.

12. 13. 14.

15. 16.

17.

5 | Permutation Groups

115

Compute each of the following. a. a21 b. ba c. ab 18. Let a5 c

19.

20. 21. 22. 23.

24. 25. 26. 27. 28. 29. 30. 31.

32.

1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 d and b 5 c d. 2 3 4 5 1 7 8 6 1 3 8 7 6 5 2 4

Write a, b, and ab as a. products of disjoint cycles, b. products of 2-cycles. Show that if H is a subgroup of Sn, then either every member of H is an even permutation or exactly half of the members are even. (This exercise is referred to in Chapter 25.) Compute the order of each member of A4. What arithmetic relationship do these orders have with the order of A4? Give two reasons why the set of odd permutations in Sn is not a subgroup. Let a and b belong to Sn. Prove that a21b21ab is an even permutation. Use Table 5.1 to compute the following. a. The centralizer of a3 5 (13)(24). b. The centralizer of a12 5 (124). How many elements of order 5 are in S7? How many elements of order 4 does S6 have? How many elements of order 2 does S6 have? Prove that (1234) is not the product of 3-cycles. Let b [ S7 and suppose b 4 5 (2143567). Find b. Let b 5 (123)(145). Write b 99 in disjoint cycle form. Find three elements s in S9 with the property that s 3 5 (157)(283)(469). What cycle is (a1a2 ? ? ? an)21? Let G be a group of permutations on a set X. Let a [ X and define stab(a) 5 {a [ G|a(a) 5 a}. We call stab(a) the stabilizer of a in G (since it consists of all members of G that leave a fixed). Prove that stab(a) is a subgroup of G. (This subgroup was introduced by Galois in 1832.) This exercise is referred to in Chapter 7. Let b 5 (1, 3, 5, 7, 9, 8, 6)(2, 4, 10). What is the smallest positive integer n for which bn 5 b25?

116

Groups

33. Let a 5 (1, 3, 5, 7, 9)(2, 4, 6)(8, 10). If am is a 5-cycle, what can you say about m? 34. Let H 5 {b [ S5|b(1) 5 1 and b(3) 5 3}. Prove that H is a subgroup of S5. How many elements are in H? Is your argument valid when 5 is replaced by any n $ 3? How many elements are in H when 5 is replaced by any n $ 3? 35. How many elements of order 5 are there in A6? 36. In S4, find a cyclic subgroup of order 4 and a noncyclic subgroup of order 4. 37. Suppose that b is a 10-cycle. For which integers i between 2 and 10 is b i also a 10-cycle? 38. In S3, find elements a and b such that |a| 5 2, |b| 5 2, and |ab| 5 3. 39. Find group elements a and b such that |a| 5 3, |b| 5 3, and |ab| 5 5. 40. Represent the symmetry group of an equilateral triangle as a group of permutations of its vertices (see Example 3). 41. Prove that Sn is non-Abelian for all n $ 3. 42. Let a and b belong to Sn. Prove that bab21 and a are both even or both odd. 43. Show that A5 has 24 elements of order 5, 20 elements of order 3, and 15 elements of order 2. (This exercise is referred to in Chapter 25.) 44. Find a cyclic subgroup of A8 that has order 4. 45. Find a noncyclic subgroup of A8 that has order 4. 46. Suppose that H is a subgroup of Sn of odd order. Prove that H is a subgroup of An. 47. Show that every element in An for n $ 3 can be expressed as a 3-cycle or a product of three cycles. 48. Show that for n $ 3, Z(Sn) 5 {e}. 49. Verify the statement made in the discussion of the Verhoeff check digit scheme based on D5 that a * s(b) 2 b * s(a) for distinct a and b. Use this to prove that s i(a) * s i11(b) 2 s i(b) * s i11(a) for all i. Prove that this implies that all transposition errors involving adjacent digits are detected. 50. Use the Verhoeff check-digit scheme based on D5 to append a check digit to 45723. 51. Prove that every element of Sn (n . 1) can be written as a product of elements of the form (1k). 52. (Indiana College Mathematics Competition) A card-shuffling machine always rearranges cards in the same way relative to the order in which they were given to it. All of the hearts arranged in order

5 | Permutation Groups

53. 54. 55. 56. 57. 58.

59.

117

from ace to king were put into the machine, and then the shuffled cards were put into the machine again to be shuffled. If the cards emerged in the order 10, 9, Q, 8, K, 3, 4, A, 5, J, 6, 2, 7, in what order were the cards after the first shuffle? Show that a permutation with odd order is an even permutation. Let G be a group. Prove or disprove that H 5 {g2 | g [ G} is a subgroup of G. (Compare with Example 5 in Chapter 3.) Determine integers n for which H 5 5a [ An |a2 5 e6 is a subgroup of An. Given that b and g are in S4 with bg 5 (1432) , gb 5 (1243) and b(1) 5 4, determine b and g. Why does the fact that the orders of the elements of A4 are 1, 2, and 3 imply that |Z(A4)| 5 1? Label the four locations of tires on an automobile with the labels 1, 2, 3, and 4, clockwise. Let a represent the operation of switching the tires in positions 1 and 3 and switching the tires in positions 2 and 4. Let b represent the operation of rotating the tires in positions 2, 3, and 4 clockwise and leaving the tire in position 1 as is. Let G be the group of all possible combinations of a and b. How many elements are in G? Shown below are four tire rotation patterns recommended by the Dunlop Tire Company. Explain how these patterns can be represented as permutations in S4 and find the smallest subgroup of S4 that contains these four patterns. Is the subgroup Abelian?

Rear Wheel Drive Vehicles

X Tires to the Driven Axle Front Wheel Drive Vehicles

FRONT

FRONT Modified X

Modified X

4 Wheel Drive Vehicles

Alternate Pattern

FRONT

FRONT

X

Normal

118

Groups

Computer Exercises Science is what we understand well enough to explain to a computer. Art is everything else we do. DONALD KNUTH, The Art of Computer Programming, 1969

Software for Computer Exercise 1 in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines whether the two permutations (1x) and (123 ? ? ? n) generate Sn for various choices of x and n (that is, whether every element of Sn can be expressed as some product of these permutations). For n 5 4, run the program for x 5 2, 3, and 4. For n 5 5, run the program for x 5 2, 3, 4, and 5. For n 5 6, run the program for x 5 2, 3, 4, 5, and 6. For n 5 8, run the program for x 5 2, 3, 4, 5, 6, 7, and 8. Conjecture a necessary and sufficient condition involving x and n for (1x) and (123 ? ? ? n) to generate Sn. 2. Use a software package (see Suggested Software on page 120) to express the following permutations in terms of the r and s given in Example 8. (For GAP, the prompt brk. means that the permutation entered is not in the group. In this situation, use Control-D to return to the main prompt. Be advised that GAP composes permutations from left to right as opposed to our method of right to left.) a. (456) b. (23) c. (12)(34) d. (12)(34)(56) 3. Repeat Example 8 for the puzzle shown here. 1 6

2

5

3 4

5 | Permutation Groups

119

References 1. J. A. Gallian, “The Mathematics of Identification Numbers,” The College Mathematics Journal 22 (1991): 194–202. 2. J. Verhoeff, Error Detecting Decimal Codes, Amsterdam: Math-ematisch Centrum, 1969.

Suggested Readings Douglas E. Ensely, “Invariants Under Actions to Amaze Your Friends,” Mathematics Magazine, Dec. 1999: 383–387. This article explains some card tricks that are based on permutation groups. Dmitry Fomin, “Getting It Together with ‘Polynominoes,’ ” Quantum, Nov./Dec. 1991: 20–23. In this article, permutation groups are used to analyze various sorts of checkerboard tiling problems. J. A. Gallian, “Error Detection Methods,” ACM Computing Surveys 28 (1996): 504–517. This article gives a comprehensive survey of error-detection methods that use check digits. This article can be downloaded at http://www.d.umn .edu/~jgallian/detection.pdf I. N. Herstein and I. Kaplansky, Matters Mathematical, New York: Chelsea, 1978. Chapter 3 of this book discusses several interesting applications of permutations to games. Douglas Hofstadter, “The Magic Cube’s Cubies Are Twiddled by Cubists and Solved by Cubemeisters,” Scientific American 244 (1981): 20–39. This article, written by a Pulitzer Prize recipient, discusses the group theory involved in the solution of the Magic (Rubik’s) Cube. In particular, permutation groups, subgroups, conjugates (elements of the form xyx21), commutators (elements of the form xyx21y21), and the “always even or always odd” theorem (Theorem 5.5) are prominently mentioned. At one point, Hofstadter says, “It is this kind of marvelously concrete illustration of an abstract notion of group theory that makes the Magic Cube one of the most amazing things ever invented for teaching mathematical ideas.” John O. Kiltinen, Oval Track & Other Permutation Puzzles & Just Enough Group Theory to Solve Them, Mathematical Association of America, Washington, D.C., 2003.

120

Groups

This book and the software that comes with it present the user with an array of computerized puzzles, plus tools to vary them in thousands of ways. The book provides the background needed to use the puzzle software to its fullest potential, and also gives the reader a gentle, not-too-technical introduction to the theory of permutation groups that is a prerequisite to a full understanding of how to solve puzzles of this type. The website http://www-instruct.nmu .edu/math_cs/kiltinen/web/mathpuzzles/ provides resources that expand upon the book. It also has news about puzzle software—modules that add functionality and fun to puzzles. Will Oakley, “Portrait of Three Puzzle Graces,” Quantum, Nov./Dec. 1991: 83–86. The author uses permutation groups to analyze solutions to the 15 puzzle, Rubik’s Cube, and Rubik’s Clock. A. White and R. Wilson, “The Hunting Group,” Mathematical Gazette 79 (1995): 5–16. This article explains how permutation groups are used in bell ringing. S. Winters, “Error-Detecting Schemes Using Dihedral Groups,” UMAP Journal 11, no. 4 (1990): 299–308. This article discusses error-detection schemes based on Dn for n odd. Schemes for both one and two check digits are analyzed.

Suggested Software GAP is free for downloading. Versions are available for Unix, Windows, and Macintosh at: http://www.gap-system.org

Augustin Cauchy You see that little young man? Well! He will supplant all of us in so far as we are mathematicians. Spoken by Lagrange to Laplace About the 11-year-old Cauchy

This stamp was issued by France in Cauchy’s honor.

AUGUSTIN LOUIS CAUCHY was born on August 21, 1789, in Paris. By the time he was 11, both Laplace and Lagrange had recognized Cauchy’s extraordinary talent for mathematics. In school he won prizes for Greek, Latin, and the humanities. At the age of 21, he was given a commission in Napoleon’s army as a civil engineer. For the next few years, Cauchy attended to his engineering duties while carrying out brilliant mathematical research on the side. In 1815, at the age of 26, Cauchy was made Professor of Mathematics at the École Polytechnique and was recognized as the leading mathematician in France. Cauchy and his contemporary Gauss were among the last mathematicians to know the whole of mathematics as known at their time, and both made important contributions to nearly

every branch, both pure and applied, as well as to physics and astronomy. Cauchy introduced a new level of rigor into mathematical analysis. We owe our contemporary notions of limit and continuity to him. He gave the first proof of the Fundamental Theorem of Calculus. Cauchy was the founder of complex function theory and a pioneer in the theory of permutation groups and determinants. His total written output of mathematics fills 24 large volumes. He wrote more than 500 research papers after the age of 50. Cauchy died at the age of 67 on May 23, 1857. For more information about Cauchy, visit: http://www–groups.dcs .st-and.ac.uk/~history/

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6 Isomorphisms

The basis for poetry and scientific discovery is the ability to comprehend the unlike in the like and the like in the unlike. JACOB BRONOWSKI

Motivation Suppose an American and a German are asked to count a handful of objects. The American says, “One, two, three, four, five, . . . ,” whereas the German says “Eins, zwei, drei, vier, fünf, . . .” Are the two doing different things? No. They are both counting the objects, but they are using different terminology to do so. Similarly, when one person says: “Two plus three is five” and another says: “Zwei und drei ist fünf,” the two are in agreement on the concept they are describing, but they are using different terminology to describe the concept. An analogous situation often occurs with groups; the same group is described with different terminology. We have seen two examples of this so far. In Chapter 1, we described the symmetries of a square in geometric terms (e.g., R90), whereas in Chapter 5 we described the same group by way of permutations of the corners. In both cases, the underlying group was the symmetries of a square. In Chapter 4, we observed that when we have a cyclic group of order n generated by a, the operation turns out to be essentially that of addition modulo n, since aras 5 ak, where k 5 (r 1 s) mod n. For example, each of U(43) and U(49) is cyclic of order 42. So, each has the form a, where aras 5 a (r 1 s) mod 42.

Definition and Examples In this chapter, we give a formal method for determining whether two groups defined in different terms are really the same. When this is the case, we say that there is an isomorphism between the two groups. This notion was first introduced by Galois about 175 years ago. The term isomorphism is derived from the Greek words isos, meaning “same” or “equal,” and morphe, meaning “form.” R. Allenby has colorfully 122

6 | Isomorphisms

123

defined an algebraist as “a person who can’t tell the difference between isomorphic systems.” Definition Group Isomorphism

An isomorphism f from a group G to a group G is a one-to-one mapping (or function) from G onto G that preserves the group operation. That is, f(ab) 5 f(a)f(b)

for all a, b in G.

If there is an isomorphism from G onto G, we say that G and G are isomorphic and write G < G.

This definition can be visualized as shown in Figure 6.1. The pairs of dashed arrows represent the group operations. G

φ b

φ

a

G

φ (b) φ (a)

φ φ (a)φ (b)

ab

Figure 6.1

It is implicit in the definition of isomorphism that isomorphic groups have the same order. It is also implicit in the definition of isomorphism that the operation on the left side of the equal sign is that of G, whereas the operation on the right side is that of G. The four cases involving ? and 1 are shown in Table 6.1. Table 6.1 G Operation

G Operation

Operation Preservation

? ? 1 1

? 1 ? 1

f(a ? b) 5 f(a) ? f(b) f(a ? b) 5 f(a) 1 f(b) f(a 1 b) 5 f(a) ? f(b) f(a 1 b) 5 f(a) 1 f(b)

There are four separate steps involved in proving that a group G is isomorphic to a group G. Step 1 “Mapping.” Define a candidate for the isomorphism; that is, define a function f from G to G. Step 2 “1–1.” Prove that f is one-to-one; that is, assume that f(a) 5 f(b) and prove that a 5 b.

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Groups

Step 3 “Onto.” Prove that f is onto; that is, for any element g in G, find an element g in G such that f(g) 5 g. Step 4 “O.P.” Prove that f is operation-preserving; that is, show that f(ab) 5 f(a)f(b) for all a and b in G. None of these steps is unfamiliar to you. The only one that may appear novel is the fourth one. It requires that one be able to obtain the same result by combining two elements and then mapping, or by mapping two elements and then combining them. Roughly speaking, this says that the two processes—operating and mapping—can be done in either order without affecting the result. This same concept arises in calculus when we say lim ( f (x) ? g(x) ) 5 lim f (x) lim g(x) xSa

xSa

xSa

or b

b

b

3 ( f 1 g) dx 5 3 f dx 1 3 g dx. a

a

a

Before going any further, let’s consider some examples. EXAMPLE 1 Let G be the real numbers under addition and let G be the positive real numbers under multiplication. Then G and G are isomorphic under the mapping f(x) 5 2x. Certainly, f is a function from G to G. To prove that it is one-to-one, suppose that 2x 5 2y. Then log2 2x 5 log2 2y, and therefore x 5 y. For “onto,” we must find for any positive real number y some real number x such that f(x) 5 y; that is, 2x 5 y. Well, solving for x gives log2 y. Finally, f(x 1 y) 5 2x1y 5 2x ? 2y 5 f(x)f(y) for all x and y in G, so that f is operation-preserving as well. EXAMPLE 2 Any infinite cyclic group is isomorphic to Z. Indeed, if a is a generator of the cyclic group, the mapping ak → k is an isomorphism. Any finite cyclic group a of order n is isomorphic to Zn under the mapping ak → k mod n. That these correspondences are functions and are one-to-one is the essence of Theorem 4.1. Obviously, the mappings are onto. That the mappings are operation-preserving follows from Exercise 11 in Chapter 0 in the finite case and from the definitions in the infinite case.

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EXAMPLE 3 The mapping from R under addition to itself given by f(x) 5 x3 is not an isomorphism. Although f is one-to-one and onto, it is not operation-preserving, since it is not true that (x 1 y)3 5 x3 1 y3 for all x and y. EXAMPLE 4 U(10) < Z4 and U(5) < Z4. To verify this, one need only observe that both U(10) and U(5) are cyclic of order 4. Then appeal to Example 2. EXAMPLE 5 U(10) ] U(12). This is a bit trickier to prove. First, note that x2 5 1 for all x in U(12). Now, suppose that f is an isomorphism from U(10) onto U(12). Then, f(9) 5 f(3 ? 3) 5 f(3)f(3) 5 1 and f(1) 5 f(1 ? 1) 5 f(1)f(1) 5 1. Thus, f(9) 5 f(1), but 9 2 1, which contradicts the assumption that f is one-to-one. EXAMPLE 6 There is no isomorphism from Q, the group of rational numbers under addition, to Q*, the group of nonzero rational numbers under multiplication. If f were such a mapping, there would be a rational number a such that f(a) 5 21. But then 21 5 f(a) 5 f( 21 a 1 21 a) 5 f( 21 a)f( 21 a) 5 [f( 21 a)]2. However, no rational number squared is 21. EXAMPLE 7 Let G 5 SL(2, R), the group of 2 3 2 real matrices with determinant 1. Let M be any 2 3 2 real matrix with determinant 1. Then we can define a mapping from G to G itself by fM(A) 5 MAM21 for all A in G. To verify that fM is an isomorphism, we carry out the four steps. Step 1 fM is a function from G to G. Here, we must show that fM(A) is indeed an element of G whenever A is. This follows from properties of determinants: det (MAM21) 5 (det M)(det A)(det M)21 5 1 ? 1 ? 121 5 1. Thus, MAM21 is in G. Step 2 fM is one-to-one. Suppose that fM(A) 5 fM(B). Then MAM21 5 MBM21 and, by left and right cancellation, A 5 B.

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Groups

Step 3 fM is onto. Let B belong to G. We must find a matrix A in G such that fM(A) 5 B. How shall we do this? If such a matrix A is to exist, it must have the property that MAM21 5 B. But this tells us exactly what A must be! For we can solve for A to obtain A 5 M21BM and verify that fM(A) 5 MAM21 5 M(M21BM)M21 5 B. Step 4 fM is operation-preserving. Let A and B belong to G. Then, fM(AB) 5 M(AB)M21 5 MA(M21M)BM21 5 (MAM21)(MBM21) 5 fM(A)fM(B). The mapping fM is called conjugation by M.

Cayley’s Theorem Our first theorem is a classic result of Cayley. An important generalization of it will be given in Chapter 25. Theorem 6.1 Cayley’s Theorem (1854) Every group is isomorphic to a group of permutations.

PROOF To prove this, let G be any group. We must find a group G of permutations that we believe is isomorphic to G. Since G is all we have to work with, we will have to use it to construct G. For any g in G, define a function Tg from G to G by Tg(x) 5 gx

for all x in G.

(In words, Tg is just multiplication by g on the left.) We leave it as an exercise (Exercise 23) to prove that Tg is a permutation on the set of elements of G. Now, let G 5 {Tg | g [ G}. Then, G is a group under the operation of function composition. To verify this, we first observe that for any g and h in G we have TgTh(x) 5 Tg(Th(x)) 5 Tg(hx) 5 g(hx) 5 (gh)x 5 Tgh(x), so that TgTh 5 Tgh. From this it follows that Te is the identity and (Tg)21 5 Tg21 (see Exercise 9). Since function composition is associative, we have verified all the conditions for G to be a group. The isomorphism f between G and G is now ready-made. For every g in G, define f(g) 5 Tg. If Tg 5 Th, then Tg(e) 5 Th(e) or ge 5 he. Thus, g 5 h and f is one-to-one. By the way G was constructed, we see that f is onto. The only condition that remains to be checked is that f is operation-preserving. To this end, let a and b belong to G. Then f(ab) 5 Tab 5 TaTb 5 f(a)f(b).

127

6 | Isomorphisms

The group G constructed above is called the left regular representation of G. EXAMPLE 8 For concreteness, let us calculate the left regular representation U(12) for U(12) 5 {1, 5, 7, 11}. Writing the permutations of U(12) in array form, we have (remember, Tx is just multiplication by x) 1 5 7 11 d, 1 5 7 11

T5 5 c

1 5 7 11 d, 7 11 1 5

T11 5 c

T1 5 c T7 5 c

1 5 7 11 d, 5 1 11 7 1 5 7 11 d. 11 7 5 1

It is instructive to compare the Cayley table for U(12) and its left regular representation U(12) . U(12)

1

5

7

11

1 5 7 11

1 5 7 11

5 1 11 7

7 11 1 5

11 7 5 1

U(12)

T1

T5

T7

T11

T1 T5 T7 T11

T1 T5 T7 T11

T5 T1 T11 T7

T7 T11 T1 T5

T11 T7 T5 T1

It should be abundantly clear from these tables that U(12) and U(12) are only notationally different. Cayley’s Theorem is important for two contrasting reasons. One is that it allows us to represent an abstract group in a concrete way. A second is that it shows that the present-day set of axioms we have adopted for a group is the correct abstraction of its much earlier predecessor—a group of permutations. Indeed, Cayley’s Theorem tells us that abstract groups are not different from permutation groups. Rather, it is the viewpoint that is different. It is this difference of viewpoint that has stimulated the tremendous progress in group theory and many other branches of mathematics in the 20th century. It is sometimes very difficult to prove or disprove, whichever the case may be, that two particular groups are isomorphic. For example, it requires somewhat sophisticated techniques to prove the surprising fact that the group of real numbers under addition is isomorphic to the group of complex numbers under addition. Likewise, it is not easy to prove the fact that the group of nonzero complex numbers under multiplication is isomorphic to the group of complex numbers with absolute value of 1 under multiplication. In geometric terms, this says that, as groups, the punctured plane and the unit circle are isomorphic. (See reference 1.)

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Groups

Properties of Isomorphisms Our next two theorems give a catalog of properties of isomorphisms and isomorphic groups. Theorem 6.2 Properties of Isomorphisms Acting on Elements Suppose that f is an isomorphism from a group G onto a group G. Then 1. f carries the identity of G to the identity of G. 2. For every integer n and for every group element a in G, f(an) 5 [f(a)]n. 3. For any elements a and b in G, a and b commute if and only if f(a) and f(b) commute. 4. G 5 a if and only if G 5 f(a). 5. |a| 5 |f(a)| for all a in G (isomorphisms preserve orders). 6. For a fixed integer k and a fixed group element b in G, the equation xk 5 b has the same number of solutions in G as does the equation xk 5 f(b) in G. 7. If G is finite, then G and G have exactly the same number of elements of every order.

PROOF We will restrict ourselves to proving only properties 1, 2, and 4, but observe that property 5 follows from properties 1 and 2, property 6 follows from property 2, and property 7 follows from property 5. For convenience, let us denote the identity in G by e and the identity in G by e. Then, since e 5 ee, we have f(e) 5 f(ee) 5 f(e)f(e). Also, because f(e) [ G, we have f(e) 5 ef(e), as well. Thus, by cancellation, e 5 f(e). This proves property 1. For positive integers, property 2 follows from the definition of an isomorphism and mathematical induction. If n is negative, then 2n is positive, and we have from property 1 and the observation about the positive integer case that e 5 f(e) 5 f(gng2n) 5 f(gn)f(g2n) 5 f(gn)(f(g))2n. Thus, multiplying both sides on the right by (f(g))n, we have (f(g))n 5 f(gn). Property 1 takes care of the case n 5 0. To prove property 4, let G 5 a and note that, by closure, f(a) # G. Because f is onto, for any element b in G, there is an element ak in G such that f(ak) 5 b. Thus, b 5 (f(a))k and so b [ f(a). This proves that G 5 f(a). Now suppose that G 5 f(a). Clearly, a # G. For any element b in G, we have f(b) [ f(a). So, for some integer k we have

6 | Isomorphisms

129

f(b) 5 (f(a))k 5 f(ak). Because f is one-to-one, b 5 ak. This proves that a 5 G. When the group operation is addition, property 2 of Theorem 6.2 is f(na) 5 nf(a); property 4 says that an isomorphism between two cyclic groups takes a generator to a generator. Property 6 is quite useful for showing that two groups are not isomorphic. Often b is picked to be the identity. For example, consider C* and R*. Because the equation x4 5 1 has four solutions in C* but only two in R*, no matter how one attempts to define an isomorphism from C* to R*, property 6 cannot hold. Theorem 6.3 Properties of Isomorphisms Acting on Groups Suppose that f is an isomorphism from a group G onto a group G. Then 1. 2. 3. 4.

f21 is an isomorphism from G onto G. G is Abelian if and only if G is Abelian. G is cyclic if and only if G is cyclic. If K is a subgroup of G, then f(K) 5 {f(k) | k [ K} is a subgroup of G.

PROOF Properties 1 and 4 are left as exercises (Exercises 21 and 22). Property 2 is a direct consequence of property 3 of Theorem 6.2. Property 3 follows from property 4 of Theorem 6.2 and property 1 of Theorem 6.3. Theorems 6.2 and 6.3 show that isomorphic groups have many properties in common. Actually, the definition is precisely formulated so that isomorphic groups have all group-theoretic properties in common. By this we mean that if two groups are isomorphic, then any property that can be expressed in the language of group theory is true for one if and only if it is true for the other. This is why algebraists speak of isomorphic groups as “equal” or “the same.” Admittedly, calling such groups equivalent, rather than the same, might be more appropriate, but we bow to long-standing tradition.

Automorphisms Certain kinds of isomorphisms are referred to so often that they have been given special names.

130

Groups

Definition Automorphism

An isomorphism from a group G onto itself is called an automorphism of G.

The isomorphism in Example 7 is an automorphism of SL(2, R). Two more examples follow. EXAMPLE 9 The function f from C to C given by f(a 1 bi) 5 a 2 bi is an automorphism of the group of complex numbers under addition. The restriction of f to C* is also an automorphism of the group of nonzero complex numbers under multiplication. (See Exercise 25.) EXAMPLE 10 Let R2 5 {(a, b) | a, b [ R}. Then f(a, b) 5 (b, a) is an automorphism of the group R2 under componentwise addition. Geometrically, f reflects each point in the plane across the line y 5 x. More generally, any reflection across a line passing through the origin or any rotation of the plane about the origin is an automorphism of R2. The isomorphism in Example 7 is a particular instance of an automorphism that arises often enough to warrant a name and notation of its own. Definition Inner Automorphism Induced by a

Let G be a group, and let a [ G. The function fa defined by fa(x) 5 axa21 for all x in G is called the inner automorphism of G induced by a.

We leave it for the reader to show that fa is actually an automorphism of G. (Use Example 7 as a model.) EXAMPLE 11 The action of the inner automorphism of D4 induced by R90 is given in the following table. x R0

fR90 → R90 x R9021 → R90R0R 90–1 5 R0

R90 → R90R90R 9021 5 R90 R180 → R90R180R 9021 5 R180 R270 → R90R270R 9021 5 R270 H

→ R90HR 9021 5 V

V

→ R90VR 9021 5 H

D

→ R90DR 9021 5 D9

D9 → R90D9R 9021 5 D

6 | Isomorphisms

131

When G is a group, we use Aut(G) to denote the set of all automorphisms of G and Inn(G) to denote the set of all inner automorphisms of G. The reason these sets are noteworthy is demonstrated by the next theorem. Theorem 6.4 Aut(G) and Inn(G) Are Groups† The set of automorphisms of a group and the set of inner automorphisms of a group are both groups under the operation of function composition.

PROOF The proof of Theorem 6.4 is left as an exercise (Exercise 15). The determination of Inn(G) is routine. If G 5 {e, a, b, c. . . .}, then Inn(G) 5 {fe, fa, fb, fc, . . .}. This latter list may have duplications, however, since fa may be equal to fb even though a 2 b (see Exercise 33). Thus, the only work involved in determining Inn(G) is deciding which distinct elements give the distinct automorphisms. On the other hand, the determination of Aut(G) is, in general, quite involved. EXAMPLE 12 Inn(D4) To determine Inn(D4), we first observe that the complete list of inner automorphisms is fR0, fR90, fR180, fR270, fH, fV, fD, and fD9. Our job is to determine the repetitions in this list. Since R180 [ Z(D4), we have fR180(x) 5 R180 xR18021 5 x, so that fR180 5 fR0. Also, fR270(x) 5 R270 xR27021 5 R90R180 xR18021R9021 5 R90 xR9021 5 fR90(x). Similarly, since H 5 R180V and D9 5 R180D, we have fH 5 fV and fD 5 fD9. This proves that the previous list can be pared down to fR0, fR90, fH, and fD. We leave it to the reader to show that these are distinct (Exercise 13). EXAMPLE 13 Aut(Z10) To compute Aut(Z10), we try to discover enough information about an element a of Aut(Z10) to determine how a must be defined. Because Z10 is so simple, this is not difficult to do. To begin with, observe that once we know a(1), we know a(k) for any k, because

†The

group Aut(G) was first studied by O. Hölder in 1893 and, independently, by E. H. Moore in 1894.

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Groups

a(k) 5 a(1 1 1 1 ? ? ? 1 1) k terms

5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 ka(1). k terms

So, we need only determine the choices for a(1) that make a an automorphism of Z10. Since property 5 of Theorem 6.2 tells us that |a(1)| 5 10, there are four candidates for a(1): a(1) 5 1;

a(1) 5 3;

a(1) 5 7;

a(1) 5 9.

To distinguish among the four possibilities, we refine our notation by denoting the mapping that sends 1 to 1 by a 1, 1 to 3 by a 3, 1 to 7 by a 7, and 1 to 9 by a 9. So the only possibilities for Aut(Z10) are a 1, a 3, a 7, and a 9. But are all these automorphisms? Clearly, a 1 is the identity. Let us check a3. Since x mod 10 5 y mod 10 implies 3x mod 10 5 3y mod 10, a3 is well defined. Moreover, because a3 (1) 5 3 is a generator of Z 10, it follows that a 3 is onto (and, by Exercise 10 in Chapter 5, it is also oneto-one). Finally, since a3(a 1 b) 5 3(a 1 b) 5 3a 1 3b 5 a3(a) 1 a3(b), we see that a 3 is operation-preserving as well. Thus, a 3 [ Aut(Z10 ). The same argument shows that a7 and a9 are also automorphisms. This gives us the elements of Aut(Z10) but not the structure. For instance, what is a 3a 3? Well, (a3a3)(1) 5 a 3(3) 5 3 ? 3 5 9 5 a9(1), so a 3 a 3 5 a 9. Similar calculations show that a 33 5 a 7 and a 34 5 a 1, so that |a3| 5 4. Thus, Aut(Z10 ) is cyclic. Actually, the following Cayley tables reveal that Aut(Z10 ) is isomorphic to U(10). U(10)

1

3

7

9

Aut(Z10)

a1

a3

a7

a9

1 3 7 9

1 3 7 9

3 9 1 7

7 1 9 3

9 7 3 1

a1 a3 a7 a9

a1 a3 a7 a9

a3 a9 a1 a7

a7 a1 a9 a3

a9 a7 a3 a1

With Example 13 as a guide, we are now ready to tackle the group Aut(Zn). The result is particularly nice, since it relates the two kinds of groups we have most frequently encountered thus far—the cyclic groups Zn and the U-groups U(n). Theorem 6.5 Aut(Zn) < U(n) For every positive integer n, Aut(Zn) is isomorphic to U(n).

6 | Isomorphisms

133

PROOF As in Example 13, any automorphism a is determined by the value of a(1), and a(1) [ U(n). Now consider the correspondence from Aut(Zn) to U(n) given by T: a → a(1). The fact that a(k) 5 ka(1) (see Example 13) implies that T is a one-to-one mapping. For if a and b belong to Aut(Zn) and a(1) 5 b(1), then a(k) 5 ka(1) 5 kb(1) 5 b(k) for all k in Zn, and therefore a 5 b. To prove that T is onto, let r [ U(n) and consider the mapping a from Zn to Zn defined by a(s) 5 sr (mod n) for all s in Zn. We leave it as an exercise to verify that a is an automorphism of Zn (see Exercise 17). Then, since T(a) 5 a(1) 5 r, T is onto U(n). Finally, we establish the fact that T is operation-preserving. Let a, b [ Aut(Zn). We then have T(ab) 5 (ab)(1) 5 a(b(1)) 5 a(1 1 1 1 ? ? ? 1 1) b(1) terms

5 a(1) 1 a(1) 1 ? ? ? 1 a(1) 5 a(1)b(1) b(1) terms

5 T(a)T(b). This completes the proof.

Exercises Being a mathematician is a bit like being a manic depressive: you spend your life alternating between giddy elation and black despair. STEVEN G. KRANTZ, A Primer of Mathematical Writing

1. Find an isomorphism from the group of integers under addition to the group of even integers under addition. 2. Find Aut(Z). 3. Let R1 be the group of positive real numbers under multiplication. Show that the mapping f(x) 5 "x is an automorphism of R1. 4. Show that U(8) is not isomorphic to U(10). 5. Show that U(8) is isomorphic to U(12). 6. Prove that the notion of group isomorphism is transitive. That is, if G, H, and K are groups and G < H and H < K, then G < K. 7. Prove that S4 is not isomorphic to D12. 8. Show that the mapping a S log10 a is an isomorphism from R+ under multiplication to R under addition. 9. In the notation of Theorem 6.1, prove that Te is the identity and that (Tg)21 5 Tg . 21

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Groups

10. Let G be a group. Prove that the mapping a(g) 5 g21 for all g in G is an automorphism if and only if G is Abelian. 11. For inner automorphisms fg , fh , and fgh, prove that fgfh 5 fgh. 12. Find two groups G and H such that G ] H, but Aut(G) < Aut(H). 13. Prove the assertion in Example 12 that the inner automorphisms fR , fR , fH, and fD of D4 are distinct. 0 90 14. Find Aut(Z6). 15. If G is a group, prove that Aut(G) and Inn(G) are groups. 16. Prove that the mapping from U(16) to itself given by x → x3 is an automorphism. What about x → x5 and x → x7? Generalize. 17. Let r [ U(n). Prove that the mapping a: Zn → Zn defined by a(s) 5 sr mod n for all s in Zn is an automorphism of Zn . (This exercise is referred to in this chapter.) 1 a 18. The group e c d ` a[ Z f is isomorphic to what familiar group? 0 1 What if Z is replaced by R? 19. If f and g are isomorphisms from the cyclic group a to some group and f(a) 5 g(a) , prove that f 5 g . 20. Suppose that f: Z 50 S Z 50 is an automorphism with f(11) 5 13. Determine a formula for f(x) . 21. Prove Property 1 of Theorem 6.3. 22. Prove Property 4 of Theorem 6.3. 23. Referring to Theorem 6.1, prove that Tg is indeed a permutation on the set G. 24. Prove or disprove that U(20) and U(24) are isomorphic. 25. Show that the mapping f(a 1 bi) 5 a 2 bi is an automorphism of the group of complex numbers under addition. Show that f preserves complex multiplication as well—that is, f(xy) 5 f(x)f(y) for all x and y in C. (This exercise is referred to in Chapter 15.) 26. Let G 5 {a 1 b"2 | a, b rational} and H5e c

a b

2b d ` a, b rational f. a

Show that G and H are isomorphic under addition. Prove that G and H are closed under multiplication. Does your isomorphism preserve multiplication as well as addition? (G and H are examples of rings—a topic we will take up in Part 3.)

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135

27. Prove that Z under addition is not isomorphic to Q under addition. 28. Prove that the quaternion group (see Exercise 4, Supplementary Exercises for Chapters 1–4) is not isomorphic to the dihedral group D4. 29. Let C be the complex numbers and M5 ec

30.

31.

32.

33.

34. 35. 36. 37.

38.

39. 40.

a b

2b d ` a, b [ R f. a

Prove that C and M are isomorphic under addition and that C* and M*, the nonzero elements of M, are isomorphic under multiplication. Let Rn 5 {(a1, a2, . . . , an) | ai [ R}. Show that the mapping f: (a1, a2, . . . , an) → (2a1, 2a2, . . . , 2an) is an automorphism of the group Rn under componentwise addition. This automorphism is called inversion. Describe the action of f geometrically. Consider the following statement: The order of a subgroup divides the order of the group. Suppose you could prove this for finite permutation groups. Would the statement then be true for all finite groups? Explain. Suppose that G is a finite Abelian group and G has no element of order 2. Show that the mapping g → g2 is an automorphism of G. Show, by example, that if G is infinite the mapping need not be an automorphism. Let G be a group and let g [ G. If z [ Z(G), show that the inner automorphism induced by g is the same as the inner automorphism induced by zg (that is, that the mappings fg and fzg are equal). If a and g are elements of a group, prove that C(a) is isomorphic to C(gag 21 ) . Suppose that g and h induce the same inner automorphism of a group G. Prove that h21g [ Z(G). Combine the results of Exercises 33 and 35 into a single “if and only if” theorem. Let a belong to a group G and let |a| be finite. Let fa be the automorphism of G given by fa(x) 5 axa21. Show that |fa| divides |a|. Exhibit an element a from a group for which 1 , |fa| , |a|. Let G 5 {0, 62, 64, 66, . . .} and H 5 {0, 63, 66, 69, . . .}. Show that G and H are isomorphic groups under addition. Does your isomorphism preserve multiplication? Generalize to the case when G 5 m and H 5 n, where m and n are integers. Suppose that f is an automorphism of D 4 such that f(R 90 ) 5 R 270 and f(V) 5 V . Determine f(D) and f(H) . In Aut(Z9), let ai denote the automorphism that sends 1 to i where gcd(i, 9) 5 1. Write a5 and a 8 as permutations of {0, 1, . . . , 8} in disjoint cycle form. [For example, a2 5 (0)(124875)(36).]

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Groups

41. Write the permutation corresponding to R90 in the left regular representation of D4 in cycle form. 42. Show that every automorphism f of the rational numbers Q under addition to itself has the form f(x) 5 xf(1). 43. Prove that Q1, the group of positive rational numbers under multiplication, is isomorphic to a proper subgroup. 44. Prove that Q, the group of rational numbers under addition, is not isomorphic to a proper subgroup of itself. 45. Prove that every automorphism of R*, the group of nonzero real numbers under multiplication, maps positive numbers to positive numbers and negative numbers to negative numbers. 46. Let G be a finite group. Show that in the disjoint cycle form of the right regular representation Tg (x) 5 xg of G each cycle has length 0g 0 . 47. Give a group-theoretic proof that Q under addition is not isomorphic to R+ under multiplication.

Reference 1. J. R. Clay, “The Punctured Plane Is Isomorphic to the Unit Circle,” Journal of Number Theory 1 (1964): 500–501.

Computer Exercise There is only one satisfying way to boot a computer. J. H. GOLDFUSS

Software for the computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software computes the order of Aut(Dn). Run the program for n 5 3, 5, 7, and 11. Make a conjecture about the order when n is prime. Run the program for n 5 4, 8, 16, and 32. Make a conjecture about the order when n is a power of 2. Run the program when n 5 6, 10, 14, and 22. Make a conjecture about the order when n is twice a prime. Run the program for n 5 9, 15, 21, and 33. Make a conjecture about the order when n is 3 times a prime. Try to deduce a general formula for the order of Aut(Dn).

Arthur Cayley Cayley is forging the weapons for future generations of physicists. PETER TAIT

ARTHUR CAYLEY was born on August 16, 1821, in England. His genius showed itself at an early age. He published his first research paper while an undergraduate of 20, and in the next year he published eight papers. While still in his early twenties, he originated the concept of n-dimensional geometry. After graduating from Trinity College, Cambridge, Cayley stayed on for three years as a tutor. At the age of 25, he began a 14year career as a lawyer. During this period, he published approximately 200 mathematical papers, many of which are now classics. In 1863, Cayley accepted the newly established Sadlerian professorship of mathematics at Cambridge University. He spent the rest of his life in that position. One of his notable accomplishments was his role in the successful effort to have women admitted to Cambridge.

Among Cayley’s many innovations in mathematics were the notions of an abstract group and a group algebra, and the matrix concept. He made major contributions to geometry and linear algebra. Cayley and his lifelong friend and collaborator J. J. Sylvester were the founders of the theory of invariants, which was later to play an important role in the theory of relativity. Cayley’s collected works comprise 13 volumes, each about 600 pages in length. He died on January 26, 1895. To find more information about Cayley, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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Cosets and Lagrange’s 7 Theorem It might be difficult, at this point, for students to see the extreme importance of this result [Lagrange’s Theorem]. As we penetrate the subject more deeply they will become more and more aware of its basic character. I. N. HERSTEIN, Topics in Algebra

Properties of Cosets In this chapter, we will prove the single most important theorem in finite group theory—Lagrange’s Theorem. But first, we introduce a new and powerful tool for analyzing a group—the notion of a coset. This notion was invented by Galois in 1830, although the term was coined by G. A. Miller in 1910. Definition Coset of H in G

Let G be a group and let H be a subset of G. For any a [ G, the set {ah | h [ H} is denoted by aH. Analogously, Ha 5 {ha | h [ H} and aHa21 5 {aha21 | h [ H}. When H is a subgroup of G, the set aH is called the left coset of H in G containing a, whereas Ha is called the right coset of H in G containing a. In this case, the element a is called the coset representative of aH (or Ha). We use |aH| to denote the number of elements in the set aH, and |Ha| to denote the number of elements in Ha.

EXAMPLE 1 Let G 5 S3 and H 5 {(1), (13)}. Then the left cosets of H in G are (1)H 5 H, (12)H 5 {(12), (12)(13)} 5 {(12), (132)} 5 (132)H, (13)H 5 {(13), (1)} 5 H, (23)H 5 {(23), (23)(13)} 5 {(23), (123)} 5 (123)H.

138

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EXAMPLE 2 Let _ 5 {R0, R180} in D4, the dihedral group of order 8. Then, R0_ 5 _, R90_ 5 {R90, R270} 5 R270_, R180_ 5 {R180, R0} 5 _, V_ 5 {V, H} 5 H_, D_ 5 {D, D9} 5 D9_. EXAMPLE 3 Let H 5 {0, 3, 6} in Z9 under addition. In the case that the group operation is addition, we use the notation a 1 H instead of aH. Then the cosets of H in Z9 are 0 1 H 5 {0, 3, 6} 5 3 1 H 5 6 1 H, 1 1 H 5 {1, 4, 7} 5 4 1 H 5 7 1 H, 2 1 H 5 {2, 5, 8} 5 5 1 H 5 8 1 H. The three preceding examples illustrate a few facts about cosets that are worthy of our attention. First, cosets are usually not subgroups. Second, aH may be the same as bH, even though a is not the same as b. Third, since in Example 1 (12)H 5 {(12), (132)} whereas H(12) 5 {(12), (123)}, aH need not be the same as Ha. These examples and observations raise many questions. When does aH 5 bH? Do aH and bH have any elements in common? When does aH 5 Ha? Which cosets are subgroups? Why are cosets important? The next lemma and theorem answer these questions. (Analogous results hold for right cosets.) Lemma Properties of Cosets Let H be a subgroup of G, and let a and b belong to G. Then, 1. 2. 3. 4. 5. 6. 7. 8.

a [ aH, aH 5 H if and only if a [ H, aH 5 bH if and only if a [ bH aH 5 bH or aH > bH 5 [, aH 5 bH if and only if a21b [ H, |aH| 5 |bH|, aH 5 Ha if and only if H 5 aHa21, aH is a subgroup of G if and only if a [ H.

PROOF 1. a 5 ae [ aH. 2. To verify property 2, we first suppose that aH 5 H. Then a 5 ae [ aH 5 H. Next, we assume that a [ H and show that aH # H

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3. 4. 5. 6.

7. 8.

and H # aH. The first inclusion follows directly from the closure of H. To show that H # aH, let h [ H. Then, since a [ H and h [ H, we know that a21h [ H. Thus, h 5 eh 5 (aa21)h 5 a(a21h) [ aH. If aH 5 bH, then a 5 ae [ aH 5 bH. Conversely, if a [ bH we have a 5 bh where h [ H, and therefore aH 5 (bh)H 5 b(hH) 5 bH. Property 4 follows directly from property 3, for if there is an element c in aH y bH, then cH 5 aH and cH 5 bH. Observe that aH 5 bH if and only if H 5 a21bH. The result now follows from property 2. To prove that |aH| 5 |bH|, it suffices to define a one-to-one mapping from aH onto bH. Obviously, the correspondence ah → bh maps aH onto bH. That it is one-to-one follows directly from the cancellation property. Note that aH 5 Ha if and only if (aH)a21 5 (Ha)a21 5 H(aa–1) 5 H—that is, if and only if aHa21 5 H. If aH is a subgroup, then it contains the identity e. Thus, aH > eH 2 0/; and, by property 4, we have aH 5 eH 5 H. Thus, from property 2, we have a [ H. Conversely, if a [ H, then, again by property 2, aH 5 H.

Although most mathematical theorems are written in symbolic form, one should also know what they say in words. In the preceding lemma, property 1 says simply that the left coset of H containing a does contain a. Property 2 says that the H “absorbs” an element if and only if the element belongs to H. Property 3 shows that a left coset of H is uniquely determined by any one of its elements. In particular, any element of a left coset can be used to represent the coset. Property 4 says—and this is very important—that two left cosets of H are either identical or disjoint. Property 5 shows how we may transfer a question about equality of left cosets of H to a question about H itself and vice versa. Property 6 says that all left cosets of H have the same size. Property 7 is analogous to property 5 in that it shows how a question about the equality of the left and right cosets of H containing a is equivalent to a question about the equality of two subgroups of G. The last property of the lemma says that H itself is the only coset of H that is a subgroup of G. Note that properties 1, 4, and 6 of the lemma guarantee that the left cosets of a subgroup H of G partition G into blocks of equal size. Indeed, we may view the cosets of H as a partitioning of G into equivalence classes under the equivalence relation defined by a , b if aH 5 bH (see Theorem 0.6). In practice, the subgroup H is often chosen so that the cosets partition the group in some highly desirable fashion. For example, if G is

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3-space R3 and H is a plane through the origin, then the coset (a, b, c) 1 H (addition is done componentwise) is the plane passing through the point (a, b, c) and parallel to H. Thus, the cosets of H constitute a partition of 3-space into planes parallel to H. If G 5 GL(2, R) and H 5 SL(2, R), then for any matrix A in G, the coset AH is the set of all 2 3 2 matrices with the same determinant as A. Thus, c

2 0 d H is the set of all 2 3 2 matrices of determinant 2 0 1

and c

1 2 d H is the set of all 2 3 2 matrices of determinant 23. 2 1

Property 4 of the lemma is useful for actually finding the distinct cosets of a subgroup. We illustrate this in the next example. EXAMPLE 4 To find the cosets of H 5 {1, 15} in G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31} we begin with H 5 {1, 15}. We can find a second coset by choosing any element not in H, say 3, as a coset representative. This gives the coset 3H 5 {3, 13}. We find our next coset by choosing a representative not already appearing in the two previously chosen cosets, say 5. This gives us the coset 5H 5 {5, 11}. We continue to form cosets by picking elements from U(32) that have not yet appeared in the previous cosets as representatives of the cosets until we have accounted for every element of U(32). We then have the complete list of all distinct cosets of H.

Lagrange’s Theorem and Consequences We are now ready to prove a theorem that has been around for more than 200 years—longer than group theory itself! (This theorem was not originally stated in group theoretic terms.) At this stage, it should come as no surprise. Theorem 7.1 Lagrange’s Theorem†: |H| Divides |G| If G is a finite group and H is a subgroup of G, then |H| divides |G|. Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|.

†Lagrange

stated his version of this theorem in 1770, but the first complete proof was given by Pietro Abbati some 30 years later.

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PROOF Let a1H, a2H, . . . , ar H denote the distinct left cosets of H in G. Then, for each a in G, we have aH 5 aiH for some i. Also, by property 1 of the lemma, a [ aH. Thus, each member of G belongs to one of the cosets aiH. In symbols, G 5 a1H < ? ? ? < ar H. Now, property 4 of the lemma shows that this union is disjoint, so that |G| 5 |a1H| 1 |a2H| 1 ? ? ? 1 |ar H|. Finally, since |aiH| 5 |H| for each i, we have |G| 5 r|H|. We pause to emphasize that Lagrange’s Theorem is a subgroup candidate criterion; that is, it provides a list of candidates for the orders of the subgroups of a group. Thus, a group of order 12 may have subgroups of order 12, 6, 4, 3, 2, 1, but no others. Warning! The converse of Lagrange’s Theorem is false. For example, a group of order 12 need not have a subgroup of order 6. We prove this in Example 5. A special name and notation have been adopted for the number of left (or right) cosets of a subgroup in a group. The index of a subgroup H in G is the number of distinct left cosets of H in G. This number is denoted by |G:H|. As an immediate consequence of the proof of Lagrange’s Theorem, we have the following useful formula for the number of distinct left (or right) cosets of H in G. Corollary 1 |G:H| 5 |G|/|H| If G is a finite group and H is a subgroup of G, then |G:H| 5 |G|/|H|.

Corollary 2 |a| Divides |G| In a finite group, the order of each element of the group divides the order of the group.

PROOF Recall that the order of an element is the order of the subgroup generated by that element. Corollary 3 Groups of Prime Order Are Cyclic A group of prime order is cyclic.

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PROOF Suppose that G has prime order. Let a [ G and a 2 e. Then, |a| divides |G| and |a| 2 1. Thus, |a| 5 |G| and the corollary follows. Corollary 4 a|G| 5 e Let G be a finite group, and let a [ G. Then, a|G| 5 e.

PROOF By Corollary 2, |G| 5 |a|k for some positive integer k. Thus, a|G| 5 a|a|k 5 ek 5 e. Corollary 5 Fermat’s Little Theorem For every integer a and every prime p, a p mod p 5 a mod p.

PROOF By the division algorithm, a 5 pm 1 r, where 0 # r , p. Thus, a mod p 5 r, and it suffices to prove that r p mod p 5 r. If r 5 0, the result is trivial, so we may assume that r [ U( p). [Recall that U(p) 5 {1, 2, . . . , p 2 1} under multiplication modulo p.] Then, by the preceding corollary, r p21 mod p 5 1 and, therefore, r p mod p 5 r. Fermat’s Little Theorem has been used in conjunction with computers to test for primality of certain numbers. One case concerned the number p 5 2257 2 1. If p is prime, then we know from Fermat’s Little Theorem that 10 p mod p 5 10 mod p and, therefore, 10 p11 mod p 5 100 mod p. Using multiple precision and a simple loop, a computer 257 was able to calculate 10 p11 mod p 5 10 2 mod p in a few seconds. The result was not 100, and so p is not prime. EXAMPLE 5 The Converse of Lagrange’s Theorem Is False† The group A4 of order 12 has no subgroups of order 6. To verify this, recall that A4 has eight elements of order 3 (a5 through a12 in the notation of Table 5.1) and suppose that H is a subgroup of order 6. Let a be any element of order 3 in A4. Since H has index 2 in A4, at most two of the cosets H, aH, and a2H are distinct. But equality of any pair of these three implies that aH 5 H, so that a [ H. (For example, if H 5 a2H, multiply on the left by a.) Thus, a subgroup of A4 of order 6 would have to contain all eight elements of order 3, which is absurd.

†The

first counterexample to the converse of Lagrange’s Theorem was given by Paolo Ruffini in 1799.

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For any prime p . 2, we know that Z2p and Dp are nonisomorphic groups of order 2p. This naturally raises the question of whether there could be other possible groups of these orders. Remarkably, with just the simple machinery available to us at this point, we can answer this question. Theorem 7.2 Classification of Groups of Order 2p Let G be a group of order 2p, where p is a prime greater than 2. Then G is isomorphic to Z2p or Dp.

PROOF We assume that G does not have an element of order 2p and show that G < Dp. We begin by first showing that G must have an element of order p. By our assumption and Lagrange’s Theorem, any nonidentity element of G must have order 2 or p. Thus, to verify our assertion, we may assume that every nonidentity element of G has order 2. In this case, we have for all a and b in the group ab 5 (ab)21 5 b21a21 5 ba, so that G is Abelian. Then, for any nonidentity elements a, b [ G with a 2 b, the set {e, a, b, ab} is closed and therefore is a subgroup of G of order 4. Since this contradicts Lagrange’s Theorem, we have proved that G must have an element of order p; call it a. Now let b be any element not in kal. Then bkal 2 kal and G 5 kal < bkal. We next claim that |b| 5 2. To see this, observe that since kal and bkal are the only two distinct cosets of kal in G, we must have b2kal 5 kal or b2kal 5 bkal. We may rule out b2kal 5 bkal, for then bkal 5 kal. On the other hand, b2kal 5 kal implies that b2 [ kal and, therefore, |b2| 5 1 or |b2| 5 p. But |b2| 5 p and |b| 2 2p imply that |b| 5 p. Then kbl 5 kb2l and therefore b [ kal, which is a contradiction. Thus, any element of G not in kal has order 2. Next consider ab. Since ab o kal, our argument above shows that |ab| 5 2. Then ab 5 (ab)21 5 b21a21 5 ba21. Moreover, this relation completely determines the multiplication table for G. [For example, a3(ba4) 5 a2(ab)a4 5 a2(ba21)a4 5 a(ab)a3 5 a(ba21)a3 5 (ab)a2 5 (ba21)a2 5 ba.] Since the multiplication table for all noncyclic groups of order 2p is uniquely determined by the relation ab 5 ba21, all noncyclic groups of order 2p must be isomorphic to each other. But of course, Dp, the dihedral group of order 2p, is one such group. As an immediate corollary, we have that S3, the symmetric group of degree 3, is isomorphic to D3.

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An Application of Cosets to Permutation Groups Lagrange’s Theorem and its corollaries dramatically demonstrate the fruitfulness of the coset concept. We next consider an application of cosets to permutation groups. Definition Stabilizer of a Point

Let G be a group of permutations of a set S. For each i in S, let stabG(i) 5 {f [ G | f(i) 5 i}. We call stabG(i) the stabilizer of i in G.

The student should verify that stabG(i) is a subgroup of G. (See Exercise 31 in Chapter 5.) Definition Orbit of a Point

Let G be a group of permutations of a set S. For each s in S, let orbG(s) 5 {f(s) | f [ G}. The set orbG(s) is a subset of S called the orbit of s under G. We use |orbG(s)| to denote the number of elements in orbG(s).

Example 6 should clarify these two definitions. EXAMPLE 6 Let G 5 {(1), (132)(465)(78), (132)(465), (123)(456), (123)(456)(78), (78)}. Then, orbG(1) 5 {1, 3, 2}, orbG(2) 5 {2, 1, 3}, orbG(4) 5 {4, 6, 5}, orbG(7) 5 {7, 8},

stabG(1) 5 {(1), (78)}, stabG(2) 5 {(1), (78)}, stabG(4) 5 {(1), (78)}, stabG(7) 5 {(1), (132)(465), (123)(456)}.

EXAMPLE 7 We may view D4 as a group of permutations of a square region. Figure 7.1(a) illustrates the orbit of the point p under D4, and Figure 7.1(b) illustrates the orbit of the point q under D4. Observe that stabD ( p) 5 {R0, D}, whereas stabD (q) 5 {R0}. 4

4

p

q (a)

(b)

Figure 7.1

The preceding two examples also illustrate the following theorem.

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Theorem 7.3 Orbit-Stabilizer Theorem Let G be a finite group of permutations of a set S. Then, for any i from S, |G| 5 |orbG (i)| |stabG(i)|.

PROOF By Lagrange’s Theorem, |G|/|stabG(i)| is the number of distinct left cosets of stabG(i) in G. Thus, it suffices to establish a oneto-one correspondence between the left cosets of stabG(i) and the elements in the orbit of i. To do this, we define a correspondence T by mapping the coset fstabG(i) to f(i) under T. To show that T is a welldefined function, we must show that astabG(i) 5 bstabG(i) implies a(i) 5 b(i). But astabG(i) 5 bstabG(i) implies a21b [ stabG(i), so that (a21b) (i) 5 i and, therefore, b(i) 5 a(i). Reversing the argument from the last step to the first step shows that T is also one-to-one. We conclude the proof by showing that T is onto orbG(i). Let j [ orbG(i). Then a(i) 5 j for some a [ G and clearly T(astabG(i)) 5 a(i) 5 j, so that T is onto. We leave as an exercise the proof of the important fact that the orbits of the elements of a set S under a group partition S (Exercise 33).

The Rotation Group of a Cube and a Soccer Ball It cannot be overemphasized that Theorem 7.3 and Lagrange’s Theorem (Theorem 7.1) are counting theorems.† They enable us to determine the numbers of elements in various sets. To see how Theorem 7.3 works, we will determine the order of the rotation group of a cube and a soccer ball. That is, we wish to find the number of essentially different ways in which we can take a cube or a soccer ball in a certain location in space, physically rotate it, and then still occupy its original location. EXAMPLE 8 Let G be the rotation group of a cube. Label the six faces of the cube 1 through 6. Since any rotation of the cube must carry each face of the cube to exactly one other face of the cube and different rotations induce different permutations of the faces, G can be viewed as a group of permutations on the set {1, 2, 3, 4, 5, 6}. Clearly, there is some rotation about a central horizontal or vertical axis that carries face number 1 to any other face, so that |orbG(1)| 5 6. Next, we consider stabG(1). Here, we are asking for all rotations of a cube that leave face number 1 where it is. Surely, there are only four such motions— rotations of 0°, 90°, 180°, and 270°—about the line perpendicular to †People

who don’t count won’t count (Anatole France).

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the face and passing through its center (see Figure 7.2). Thus, by Theorem 7.3, |G| 5 |orbG(1)| |stabG(1)| 5 6 ? 4 5 24.

Figure 7.2 Axis of rotation of a cube.

Now that we know how many rotations a cube has, it is simple to determine the actual structure of the rotation group of a cube. Recall that S4 is the symmetric group of degree 4. Theorem 7.4 The Rotation Group of a Cube The group of rotations of a cube is isomorphic to S4.

PROOF Since the group of rotations of a cube has the same order as S4, we need only prove that the group of rotations is isomorphic to a subgroup of S4. To this end, observe that a cube has four diagonals and that the rotation group induces a group of permutations on the four diagonals. But we must be careful not to assume that different rotations correspond to different permutations. To see that this is so, all we need do is show that all 24 permutations of the diagonals arise from rotations. Labeling the consecutive diagonals 1, 2, 3, and 4, it is obvious that there is a 90° rotation that yields the permutation a 5 (1234); another 90° rotation about an axis perpendicular to our first axis yields the permutation b 5 (1423). See Figure 7.3. So, the group of permutations induced by the rotations contains the eight-element subgroup {e, a, a2, a3, b2, b2a, b2a2, b2a3} (see Exercise 37) and ab, which has order 3. Clearly, then, the rotations yield all 24 permutations since the order of the rotation group must be divisible by both 8 and 3. EXAMPLE 9 A traditional soccer ball has 20 faces that are regular hexagons and 12 faces that are regular pentagons. (The technical term for this solid is truncated icosahedron.) To determine the number of rotational symmetries of a soccer ball using Theorem 7.3, we may choose

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2

1

2

1

4

3

4

3

3

4

3

4 2

1

2

1

α = (1234)

β = (1423)

Figure 7.3

our set S to be the 20 hexagons or the 12 pentagons. Let us say that S is the set of 12 pentagons. Since any pentagon can be carried to any other pentagon by some rotation, the orbit of any pentagon is S. Also, there are five rotations that fix (stabilize) any particular pentagon. Thus, by the Orbit-Stabilizer Theorem, there are 12 ? 5 5 60 rotational symmetries. (In case you are interested, the rotation group of a soccer ball is isomorphic to A5.)

In 1985, chemists Robert Curl, Richard Smalley, and Harold Kroto caused tremendous excitement in the scientific community when they created a new form of carbon by using a laser beam to vaporize graphite. The structure of the new molecule is composed of 60 carbon atoms arranged in the shape of a soccer ball! Because the shape of the new molecule reminded them of the dome structures built by the architect R. Buckminster Fuller, Curl, Smalley, and Kroto named their discovery “buckyballs.” Buckyballs are the roundest, most symmetrical large molecules known. Group theory has been particularly useful in illuminating the properties of buckyballs, since the absorption spectrum of a molecule depends on its symmetries and chemists classify various molecular states

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according to their symmetry properties. The buckyball discovery spurred a revolution in carbon chemistry. In 1996, Curl, Smalley, and Kroto received the Nobel Prize in chemistry for their discovery.

Exercises I don’t know, Marge. Trying is the first step towards failure. HOMER SIMPSON

1. Let H 5 {(1), (12)(34), (13)(24), (14)(23)}. Find the left cosets of H in A4 (see Table 5.1 on page 107). 2. Let H be as in Exercise 1. How many left cosets of H in S4 are there? (Determine this without listing them.) 3. Let H 5 {0, 63, 66, 69, . . .}. Find all the left cosets of H in Z. 4. Rewrite the condition a21b [ H given in property 5 of the lemma on page 139 in additive notation. Assume that the group is Abelian. 5. Let H be as in Exercise 3. Use Exercise 4 to decide whether or not the following cosets of H are the same. a. 11 1 H and 17 1 H b. 21 1 H and 5 1 H c. 7 1 H and 23 1 H 6. Let n be a positive integer. Let H 5 {0, 6n, 62n, 63n, . . .}. Find all left cosets of H in Z. How many are there? 7. Find all of the left cosets of {1, 11} in U(30). 8. Suppose that a has order 15. Find all of the left cosets of a5 in a. 9. Let |a| 5 30. How many left cosets of a4 in a are there? List them. 10. Let a and b be nonidentity elements of different orders in a group G of order 155. Prove that the only subgroup of G that contains a and b is G itself. 11. Let H be a subgroup of R*, the group of nonzero real numbers under multiplication. If R+ # H # R*, prove that H 5 R+ or H 5 R*. 12. Let C* be the group of nonzero complex numbers under multiplication and let H 5 {a + bi [ C*| a2 + b2 5 1}. Give a geometric description of the coset (3 + 4i)H. Give a geometric description of the coset (c + di)H. 13. Let G be a group of order 60. What are the possible orders for the subgroups of G? 14. Suppose that K is a proper subgroup of H and H is a proper subgroup of G. If |K| 5 42 and |G| 5 420, what are the possible orders of H?

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15. Let G be a group with |G| 5 pq, where p and q are prime. Prove that every proper subgroup of G is cyclic. 16. Recall that, for any integer n greater than 1, f(n) denotes the number of positive integers less than n and relatively prime to n. Prove that if a is any integer relatively prime to n, then af(n) mod n 5 1. 17. Compute 515 mod 7 and 713 mod 11. 18. Use Corollary 2 of Lagrange’s Theorem (Theorem 7.1) to prove that the order of U(n) is even when n . 2. 19. Suppose G is a finite group of order n and m is relatively prime to n. If g [ G and gm 5 e, prove that g 5 e. 20. Suppose H and K are subgroups of a group G. If |H| 5 12 and |K| 5 35, find |H > K|. Generalize. 21. Suppose that H is a subgroup of S4 and that H contains (12) and (234.) Prove that H 5 S4. 22. Suppose that H and K are subgroups of G and there are elements a and b in G such that aH 8 bK. Prove that H 8 K. 23. Suppose that G is an Abelian group with an odd number of elements. Show that the product of all of the elements of G is the identity. 24. Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime. (Do not assume at the outset that G is finite.) 25. Let |G| 5 15. If G has only one subgroup of order 3 and only one of order 5, prove that G is cyclic. Generalize to |G| 5 pq, where p and q are prime. 26. Let G be a group of order 25. Prove that G is cyclic or g5 5 e for all g in G. 27. Let |G| 5 33. What are the possible orders for the elements of G? Show that G must have an element of order 3. 28. Let |G| 5 8. Show that G must have an element of order 2. 29. Can a group of order 55 have exactly 20 elements of order 11? Give a reason for your answer. 30. Determine all finite subgroups of C*, the group of nonzero complex numbers under multiplication. 31. Let H and K be subgroups of a finite group G with H # K # G. Prove that |G:H| 5 |G:K| |K:H|. 32. Show that Q, the group of rational numbers under addition, has no proper subgroup of finite index.

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33. Let G be a group of permutations of a set S. Prove that the orbits of the members of S constitute a partition of S. (This exercise is referred to in this chapter and in Chapter 29.) 34. Prove that every subgroup of Dn of odd order is cyclic. 35. Let G 5 {(1), (12)(34), (1234)(56), (13)(24), (1432)(56), (56)(13), (14)(23), (24)(56)}. a. Find the stabilizer of 1 and the orbit of 1. b. Find the stabilizer of 3 and the orbit of 3. c. Find the stabilizer of 5 and the orbit of 5. 36. Let G be a group of order pn where p is prime. Prove that the center of G cannot have order pn21. 37. Prove that the eight-element set in the proof of Theorem 7.4 is a group. 38. Prove that a group of order 12 must have an element of order 2. 39. Suppose that a group contains elements of orders 1 through 10. What is the minimum possible order of the group? 40. Let G be a finite Abelian group and let n be a positive integer that is relatively prime to |G|. Show that the mapping a → an is an automorphism of G. 41. Show that in a group G of odd order, the equation x2 5 a has a unique solution for all a in G. 42. Let G be a group of order pqr, where p, q, and r are distinct primes. If H and K are subgroups of G with |H| 5 pq and |K| 5 qr, prove that |H > K| 5 q. 43. Let G 5 GL(2, R) and H 5 SL(2, R). Let A [ G and suppose that det A 5 2. Prove that AH is the set of all 2 3 2 matrices in G that have determinant 2. 44. Let G be the group of rotations of a plane about a point P in the plane. Thinking of G as a group of permutations of the plane, describe the orbit of a point Q in the plane. (This is the motivation for the name “orbit.”) 45. Let G be the rotation group of a cube. Label the faces of the cube 1 through 6, and let H be the subgroup of elements of G that carry face 1 to itself. If s is a rotation that carries face 2 to face 1, give a physical description of the coset Hs. 46. The group D4 acts as a group of permutations of the square regions shown on the following page. (The axes of symmetry are drawn for reference purposes.) For each square region, locate the points in

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the orbit of the indicated point under D4. In each case, determine the stabilizer of the indicated point.

47. Let G 5 GL(2, R), the group of 2 3 2 matrices over R with nonzero determinant. Let H be the subgroup of matrices of determinant 61. If a, b [ G and aH 5 bH, what can be said about det (a) and det (b)? Prove or disprove the converse. 48. Calculate the orders of the following (refer to Figure 27.5 for illustrations): a. The group of rotations of a regular tetrahedron (a solid with four congruent equilateral triangles as faces) b. The group of rotations of a regular octahedron (a solid with eight congruent equilateral triangles as faces) c. The group of rotations of a regular dodecahedron (a solid with 12 congruent regular pentagons as faces) d. The group of rotations of a regular icosahedron (a solid with 20 congruent equilateral triangles as faces) 49. If G is a finite group with fewer than 100 elements and G has subgroups of orders 10 and 25, what is the order of G? 50. A soccer ball has 20 faces that are regular hexagons and 12 faces that are regular pentagons. Use Theorem 7.3 to explain why a soccer ball cannot have a 60° rotational symmetry about a line through the centers of two opposite hexagonal faces.

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Computer Exercise In the fields of observation chance favors only the prepared mind. LOUIS PASTEUR

Software for the computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines when Zn is the only group of order n in the case that n 5 pq where p and q are distinct primes. Run the software for n 5 3 ? 5, 3 ? 7, 3 ? 11, 3 ? 13, 3 ? 17, 3 ? 31, 5 ? 7, 5 ? 11, 5 ? 13, 5 ? 17, 5 ? 31, 7 ? 11, 7 ? 13, 7 ? 17, 7 ? 19, and 7 ? 43. Conjecture a necessary and sufficient condition about p and q for Zpq to be the only group of order pq, where p and q are distinct primes.

Joseph Lagrange Lagrange is the Lofty Pyramid of the Mathematical Sciences. NAPOLEON BONAPARTE

JOSEPH LOUIS LAGRANGE was born in Italy of French ancestry on January 25, 1736. He became captivated by mathematics at an early age when he read an essay by Halley on Newton’s calculus. At the age of 19, he became a professor of mathematics at the Royal Artillery School in Turin. Lagrange made significant contributions to many branches of mathematics and physics, among them the theory of numbers, the theory of equations, ordinary and partial differential equations, the calculus of variations, analytic geometry, fluid dynamics, and celestial mechanics. His methods for solving third- and fourth-degree polynomial equations by radicals laid the groundwork for the group-theoretic approach to solving polynomials taken by Galois. Lagrange was a very careful writer with a clear and elegant style. At the age of 40, Lagrange was appointed Head of the Berlin Academy, succeeding Euler. In offering this appointment, Frederick the Great proclaimed that the “greatest king in Europe” ought to have the “greatest mathematician in Europe” at his court. In 1787, Lagrange was invited to Paris by Louis XVI and became a good friend of the king and his wife, Marie Antoinette. In 1793, Lagrange headed a commission, which included Laplace and Lavoisier, to devise a new system

154

This stamp was issued by France in Lagrange’s honor in 1958.

of weights and measures. Out of this came the metric system. Late in his life he was made a count by Napoleon. Lagrange died on April 10, 1813. To find more information about Lagrange, visit: http://www-groups.dcs .st-and.ac.uk/~history/

External Direct 8 Products The universe is an enormous direct product of representations of symmetry groups. STEVEN WEINBERG†

Definition and Examples In this chapter, we show how to piece together groups to make larger groups. In Chapter 9, we will show that we can often start with one large group and decompose it into a product of smaller groups in much the same way as a composite positive integer can be broken down into a product of primes. These methods will later be used to give us a simple way to construct all finite Abelian groups. Definition External Direct Product

Let G1, G2, . . . , Gn be a finite collection of groups. The external direct product of G1, G2, . . . , Gn, written as G1 % G2 % ? ? ? % Gn, is the set of all n-tuples for which the ith component is an element of Gi and the operation is componentwise. In symbols,

G1 % G2 % ? ? ? % Gn 5 {(g1, g2, . . . , gn) | gi [ Gi}, where (g1, g2, . . . , gn)(g9, 1 g9, 2 . . . , g9) n is defined to be (g1g9, 1 g2g9, . . . , g g 9). It is understood that each product gig9i is performed 2 n n with the operation of Gi. We leave it to the reader to show that the external direct product of groups is itself a group (Exercise 1). This construction is not new to students who have had linear algebra or physics. Indeed, R2 5 R % R and R3 5 R % R % R—the operation being componentwise addition. Of course, there is also scalar multiplication, but †Weinberg

received the 1979 Nobel Prize in physics with Sheldon Glashow and Abdus Salam for their construction of a single theory incorporating weak and electromagnetic interactions. 155

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we ignore this for the time being, since we are interested only in the group structure at this point. EXAMPLE 1 U(8) % U(10) 5 {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3), (3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9), (7, 1),(7, 3), (7, 7), (7, 9)}. The product (3, 7)(7, 9) 5 (5, 3), since the first components are combined by multiplication modulo 8, whereas the second components are combined by multiplication modulo 10. EXAMPLE 2 Z2 % Z3 5 {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)}. Clearly, this is an Abelian group of order 6. Is this group related to another Abelian group of order 6 that we know, namely, Z6? Consider the subgroup of Z2 % Z3 generated by (1, 1). Since the operation in each component is addition, we have (1, 1) 5 (1, 1), 2(1, 1) 5 (0, 2), 3(1, 1) 5 (1, 0), 4(1, 1) 5 (0, 1), 5(1, 1) 5 (1, 2), and 6(1, 1) 5 (0, 0). Hence Z2 % Z3 is cyclic. It follows that Z2 % Z3 is isomorphic to Z6. In Theorem 7.2 we classified the groups of order 2p where p is an odd prime. Now that we have defined Z2 % Z2, it is easy to classify the groups of order 4. EXAMPLE 3 Classification of Groups of Order 4 A group of order 4 is isomorphic to Z4 or Z2 % Z2. To verify this, let G 5 {e, a, b, ab}. If G is not cyclic, then it follows from Lagrange’s Theorem that |a| 5 |b| 5 |ab| 5 2. Then the mapping e S (0, 0), a S (1, 0), b S (0, 1), and ab S (1, 1) is an isomorphism from G onto Z2 % Z2. We see from Examples 2 and 3 that in some cases Zm % Zn is isomorphic to Zmn and in some cases it is not. Theorem 8.2 provides a simple characterization for when the isomorphism holds.

Properties of External Direct Products Our first theorem gives a simple method for computing the order of an element in a direct product in terms of the orders of the component pieces.

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Theorem 8.1 Order of an Element in a Direct Product The order of an element in a direct product of a finite number of finite groups is the least common multiple of the orders of the components of the element. In symbols,

|(g1, g2, . . . , gn)| 5 lcm(|g1|, |g2|, . . . , |gn|). PROOF Denote the identity of Gi by ei. Let s 5 lcm(|g1|, |g2|, . . . , |gn|) and t 5|(g1, g2, . . . , gn)|. Because s is a multiple of each |gi| implies that (g1, g2, . . . , gn)s 5 (g1s, g s2, . . . , gns) 5 (e1, e2, . . . , en), we know that t # s. On the other hand, from (g1t , g t2, . . . , gnt ) 5 (g1, g 2, . . . , gn)t 5(e1, e2, . . . , en) we see that t is a common multiple of |g1|, |g2|, . . . , |gn|. Thus, s # t. The next two examples are applications of Theorem 8.1. EXAMPLE 4 We determine the number of elements of order 5 in Z25 % Z5. By Theorem 8.1, we may count the number of elements (a, b) in Z25 % Z5 with the property that 5 5 |(a, b)| 5 lcm(|a|, |b|). Clearly this requires that either |a| 5 5 and |b| 5 1 or 5, or |b| 5 5 and |a| 5 1 or 5. We consider two mutually exclusive cases. Case 1 |a| 5 5 and |b| 5 1 or 5. Here there are four choices for a (namely, 5, 10, 15, and 20) and five choices for b. This gives 20 elements of order 5. Case 2 |a| 5 1 and |b| 5 5. This time there is one choice for a and four choices for b, so we obtain four more elements of order 5. Thus, Z25 % Z5 has 24 elements of order 5. EXAMPLE 5 We determine the number of cyclic subgroups of order 10 in Z100 % Z25. We begin by counting the number of elements (a, b) of order 10. Case 1 |a| 5 10 and |b| 5 1 or 5. Since Z100 has a unique cyclic subgroup of order 10 and any cyclic group of order 10 has four generators (Theorem 4.4), there are four choices for a. Similarly, there are five choices for b. This gives 20 possibilities for (a, b). Case 2 |a| 5 2 and |b| 5 5. Since any finite cyclic group of even order has a unique subgroup of order 2 (Theorem 4.4), there is only one choice for a. Obviously, there are four choices for b. So, this case yields four more possibilities for (a, b).

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Thus, Z100 % Z25 has 24 elements of order 10. Because each cyclic subgroup of order 10 has four elements of order 10 and no two of the cyclic subgroups can have an element of order 10 in common, there must be 24/4 5 6 cyclic subgroups of order 10. (This method is analogous to determining the number of sheep in a flock by counting legs and dividing by 4.) The direct product notation is convenient for specifying certain subgroups of a direct product. EXAMPLE 6 For each divisor r of m and s of n the group Zm % Zn has a subgroup isomorphic to Zr % Zs (see Exercise 17). To find a subgroup of say Z30 % Z12 isomorphic to Z6 % Z4 we observe that 5 is a subgroup of Z30 of order 6 and 3 is a subgroup of Z12 of order 4, so 5 % 3 is the desired subgroup. The next theorem and its first corollary characterize those direct products of cyclic groups that are themselves cyclic. Theorem 8.2 Criterion for G % H to be Cyclic Let G and H be finite cyclic groups. Then G % H is cyclic if and only if |G| and |H| are relatively prime.

PROOF Let |G| 5 m and |H| 5 n, so that |G % H| 5 mn. To prove the first half of the theorem, we assume G % H is cyclic and show that m and n are relatively prime. Suppose that gcd(m, n) 5 d and (g, h) is a generator of G % H. Since (g, h)mn/d 5 ((gm)n/d, (hn)m/d ) 5 (e, e), we have mn 5 |(g, h)| # mn/d. Thus, d 5 1. To prove the other half of the theorem, let G 5 g and H 5 h and suppose gcd(m, n) 5 1. Then, |(g, h)| 5 lcm(m, n) 5 mn 5 |G % H|, so that (g, h) is a generator of G % H. As a consequence of Theorem 8.2 and an induction argument, we obtain the following extension of Theorem 8.2. Corollary 1 Criterion for G1 % G2 % ? ? ? % Gn to Be Cyclic An external direct product G1 % G2 % ? ? ? % Gn of a finite number of finite cyclic groups is cyclic if and only if |Gi| and |Gj| are relatively prime when i 2 j.

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Corollary 2 Criterion for Zn1n2 . . . nk < Zn1 % Zn2 % . . . % Znk Let m 5 n1 n2 ? ? ? nk. Then Zm is isomorphic to Zn1 % Zn2 % ? ? ? % Znk if and only if ni and nj are relatively prime when i 2 j.

By using the results above in an iterative fashion, one can express the same group (up to isomorphism) in many different forms. For example, we have Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5 < Z2 % Z30. Similarly, Z2 % Z2 % Z3 % Z5 < Z2 % Z6 % Z5 < Z2 % Z3 % Z2 % Z5 < Z6 % Z10. Thus, Z2 % Z30 < Z6 % Z10. Note, however, that Z2 % Z30 ] Z60.

The Group of Units Modulo n As An External Direct Product The U-groups provide a convenient way to illustrate the preceding ideas. We first introduce some notation. If k is a divisor of n, let Uk(n) 5 {x [ U(n)| x mod k 5 1}. For example, U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92}. It can be readily shown that Uk(n) is indeed a subgroup of U(n). (See Exercise 17 in Chapter 3.) Theorem 8.3 U(n) as an External Direct Product Suppose s and t are relatively prime. Then U(st) is isomorphic to the external direct product of U(s) and U(t). In short, U(st) < U(s) % U(t). Moreover, Us(st) is isomorphic to U(t) and Ut(st) is isomorphic to U(s).

PROOF An isomorphism from U(st) to U(s) % U(t) is x → (x mod s, x mod t); an isomorphism from Us(st) to U(t) is x → x mod t; an isomorphism from Ut (st) to U(s) is x → x mod s. We leave the verification that these mappings are operation-preserving, one-to-one, and onto to the reader. (See Exercises 11, 17, and 19 in Chapter 0; see also [1].) As a consequence of Theorem 8.3, we have the following result.

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Corollary Let m 5 n1n2 ? ? ? nk , where gcd(ni , nj ) 5 1 for i 2 j. Then, U(m) < U(n1) % U(n2) % ? ? ? % U(nk).

To see how these results work, let’s apply them to U(105). We obtain U(105) < U(7) % U(15) U(105) < U(21) % U(5) U(105) < U(3) % U(5) % U(7). Moreover, U(7) < U15(105) 5 {1, 16, 31, 46, 61, 76} U(15) < U7(105) 5 {1, 8, 22, 29, 43, 64, 71, 92} U(21) < U5(105) 5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101} U(5) < U21(105) 5 {1, 22, 43, 64} U(3) < U35(105) 5 {1, 71}. Among all groups, surely the cyclic groups Zn have the simplest structures and, at the same time, are the easiest groups with which to compute. Direct products of groups of the form Zn are only slightly more complicated in structure and computability. Because of this, algebraists endeavor to describe a finite Abelian group as such a direct product. Indeed, we shall soon see that every finite Abelian group can be so represented. With this goal in mind, let us reexamine the U-groups. Using the corollary to Theorem 8.3 and the facts (see [2, p. 93]), first proved by Carl Gauss in 1801, that U(2) < {0},

U(4) < Z2,

U(2n) < Z2 % Z2n22

for n $ 3,

and U( pn) < Zpn2pn21

for p an odd prime,

we now can write any U-group as an external direct product of cyclic groups. For example, U(105) 5 U(3 ? 5 ? 7) < U(3) % U(5) % U(7) and

< Z2 % Z4 % Z6 U(720) 5 U(16 ? 9 ? 5) < U(16) % U(9) % U(5) < Z2 % Z4 % Z6 % Z4.

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What is the advantage of expressing a group in this form? Well, for one thing, we immediately see that the orders of the elements U(720) can only be 1, 2, 3, 4, 6, and 12. This follows from the observations that an element from Z2 % Z4 % Z6 % Z4 has the form (a, b, c, d), where |a| 5 1 or 2, |b| 5 1, 2, or 4, |c| 5 1, 2, 3, or 6, and |d| 5 1, 2, or 4, and that |(a, b, c, d)| 5 lcm(|a|, |b|, |c|, |d|). For another thing, we can readily determine the number of elements of order 12, say, that U(720) has. Because U(720) is isomorphic to Z2 % Z4 % Z6 % Z4, it suffices to calculate the number of elements of order 12 in Z2 % Z4 % Z6 % Z4. But this is easy. By Theorem 8.1, an element (a, b, c, d) has order 12 if and only if lcm(|a|, |b|, |c|, |d|) 5 12. Since |a| 5 1 or 2, it does not matter how a is chosen. So, how can we have lcm(|b|, |c|, |d|) 5 12? One way is to have |b| 5 4, |c| 5 3 or 6, and d arbitrary. By Theorem 4.4, there are two choices for b, four choices for c, and four choices for d. So, in this case, we have 2 ? 4 ? 4 5 32 choices. The only other way to have lcm(|b|, |c|, |d|) 5 12 is for |d| 5 4, |c| 5 3 or 6, and |b| 5 1 or 2 (we exclude |b| 5 4, since this was already accounted for). This gives 2 ? 4 ? 2 5 16 new choices. Finally, since a can be either of the two elements in Z2, we have a total of 2(32 1 16) 5 96 elements of order 12. These calculations tell us more. Since Aut(Z720) is isomorphic to U(720), we also know that there are 96 automorphisms of Z720 of order 12. Imagine trying to deduce this information directly from U(720) or, worse yet, from Aut(Z720)! These results beautifully illustrate the advantage of being able to represent a finite Abelian group as a direct product of cyclic groups. They also show the value of our theorems about Aut(Zn) and U(n). After all, theorems are laborsaving devices. If you want to convince yourself of this, try to prove directly from the definitions that Aut(Z720) has exactly 96 elements of order 12.

Applications We conclude this chapter with five applications of the material presented here—three to cryptography, the science of sending and deciphering secret messages, one to genetics, and one to electric circuits. Data Security Because computers are built from two-state electronic components, it is natural to represent information as strings of 0s and 1s called binary strings. A binary string of length n can naturally be thought of as an element of Z2 % Z2 % ? ? ? % Z2 (n copies) where the parentheses and the commas have been deleted. Thus the binary string

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11000110 corresponds to the element (1, 1, 0, 0, 0, 1, 1, 0) in Z2 % Z2 % Z2 % Z2 % Z2 % Z2 % Z2 % Z2. Similarly, two binary strings a1a2 ? ? ? an and b1b2 ? ? ? bn are added componentwise modulo 2 just as their corresponding elements in Z2 % Z2 % ? ? ? % Z2 are. For example, 11000111 1 01110110 5 10110001 and 10011100 1 10011100 5 00000000. The fact that the sum of two binary sequences a1a2 ? ? ? an 1 b1b2 ? ? ? bn 5 00 ? ? ? 0 if and only if the sequences are identical is the basis for a data security system used to protect internet transactions. Suppose that you want to purchase a compact disc from www .Amazon.com. Need you be concerned that a hacker will intercept your credit-card number during the transaction? As you might expect, your credit-card number is sent to Amazon in a way that protects the data. We explain one way to send credit-card numbers over the Web securely. When you place an order with Amazon the company sends your computer a randomly generated string of 0’s and 1’s called a key. This key has the same length as the binary string corresponding to your credit-card number and the two strings are added (think of this process as “locking” the data). The resulting sum is then transmitted to Amazon. Amazon in turn adds the same key to the received string which then produces the original string corresponding to your creditcard number (adding the key a second time “unlocks” the data). To illustrate the idea, say you want to send an eight-digit binary string such as s 5 10101100 to Amazon (actual credit-card numbers have very long strings) and Amazon sends your computer the key k 5 00111101. Your computer returns the string s 1 k 5 10101100 1 00111101 5 10010001 to Amazon, and Amazon adds k to this string to get 10010001 1 00111101 5 10101100, which is the string representing your credit-card number. If someone intercepts the number s 1 k 5 10010001 during transmission it is no value without knowing k. The method is secure because the key sent by Amazon is randomly generated and used only one time. You can tell when you are using an encryption scheme on a web transaction by looking to see if the web address begins with “https” rather than the customary “http.” You will also see a small padlock in the status bar at the bottom of the browser window. Application to Public Key Cryptography In the mid-1970s, Ronald Rivest, Adi Shamir, and Leonard Adleman devised an ingenious method that permits each person who is to receive a secret message to tell publicly how to scramble messages

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sent to him or her. And even though the method used to scramble the message is known publicly, only the person for whom it is intended will be able to unscramble the message. The idea is based on the fact that there exist efficient methods for finding very large prime numbers (say about 100 digits long) and for multiplying large numbers, but no one knows an efficient algorithm for factoring large integers (say about 200 digits long). So, the person who is to receive the message chooses a pair of large primes p and q and chooses an integer r with 1 , r , m, where m 5 lcm( p 2 1, q 2 1), such that r is relatively prime to m (any such r will do). This person calculates n 5 pq and announces that a message M is to be sent to him or her publicly as Mr mod n. Although r, n, and Mr are available to everyone, only the person who knows how to factor n as pq will be able to decipher the message. To present a simple example that nevertheless illustrates the principal features of the method, say we wish to send the message “YES.” We convert the message into a string of digits by replacing A by 01, B by 02, . . . , Z by 26, and a blank by 00. So, the message YES becomes 250519. To keep the numbers involved from becoming too unwieldy, we send the message in blocks of four digits and fill in with blanks when needed. Thus, the message YES is represented by the two blocks 2505 and 1900. The person to whom the message is to be sent has picked two primes p and q, say p 5 37 and q 5 73 (in actual practice, p and q would have 100 or so digits), and a number r that has no prime divisors in common with lcm( p 2 1, q 2 1) 5 72, say r 5 5, and has published n 5 37 ? 73 5 2701 and r 5 5 in a public directory. We will send the “scrambled” numbers (2505)5 mod 2701 and (1900)5 mod 2701 rather than 2505 and 1900, and the receiver will unscramble them. We show the work involved for us and the receiver only for the block 2505. The arithmetic involved in computing these numbers is simplified as follows: 2505 mod 2701 5 2505 (2505)2 mod 2701 5 602 (2505)4 mod 2701 5 (602)(602) mod 2701 5 470. So, (2505)5 mod 2701 5 (2505)(470) mod 2701 5 2415.† †To

determine 25052 mod 2701 with a calculator, enter 2505 3 2505 to obtain 62750025, then divide 6275025 by 2701 to obtain 2323.2228. Finally, enter 6275025 2 (2323 3 2701) to obtain 602. Provided that the numbers are not too large, the Google search engine at http://www.google.com will do modular arithmetic. For example, entering 2505^2 mod 2701 in the search box yields 602. Be careful, however, because entering 2505^5 mod 2701 computes the wrong value since 2505 5 is too large. Instead, we can use Google to compute smaller powers such as 25053 mod 2701 (which yields 852) and 25052 mod 2701 and then compute (852 3 602) mod 2701 5 2415.

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Thus, the number 2415 is sent to the receiver. Now the receiver must take this number and convert it back to 2505. To do so, the receiver takes the two factors of 2701, p 5 37 and q 5 73, and calculates the least common multiple of p 2 1 5 36 and q 2 1 5 72, which is 72. (This is where the knowledge of p and q is necessary.) Next, the receiver must find s 5 r21 in U(72)—that is, solve the equation 5 ? s 5 1 mod 72. This number is 29. (There is a simple algorithm for finding this number.) Then the receiver takes the number received, 2415, and calculates (2415)29 mod 2701. This calculation can be simplified as follows: 2415 mod 2701 5 2415 (2415)2 mod 2701 5 766 (2415)4 mod 2701 5 (766)2 mod 2701 5 639 (2415)8 mod 2701 5 (639)2 mod 2701 5 470 (2415)16 mod 2701 5 (470)2 mod 2701 5 2119 So, (2415)29 mod 2701 5 (2415)16(2415)8(2415)4(2415) mod 2701 5 (2119)(470)(639)(2415) mod 2701 5 ((2119)(470) mod 2701 3 (639)(2415) mod 2701) mod 2701 5 (1962)(914) mod 2701 5 2505. [We compute the product (2119)(470)(639)(2415) in two stages so that we may use a hand calculator.] Thus the receiver correctly determines the code for “YE.” On the other hand, without knowing how pq factors, one cannot find the modulus (in our case, 72) that is needed to determine the intended message. The procedure just described is called the RSA public key encryption scheme in honor of the three people (Rivest, Shamir, and Adleman) who discovered the method. It is widely used in conjunction with web servers and browsers, e-mail programs, remote login sessions, and electronic financial transactions. The algorithm is summarized below. Receiver

1. Pick very large primes p and q and compute n 5 pq. 2. Compute the least common multiple of p 2 1 and q 2 1; let us call it m. 3. Pick r relatively prime to m. 4. Find s such that rs mod m 5 1. 5. Publicly announce n and r. Sender

1. Convert the message to a string of digits. (In practice, the ASCII code is used.) 2. Break up the message into uniform blocks of digits; call them M1, M2, . . . , Mk.

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3. Check to see that the greatest common divisor of each Mi and n is 1. If not, n can be factored and our code is broken. (In practice, the primes p and q are so large that they exceed all Mi, so this step may be omitted.) 4. Calculate and send Ri 5 Mi r mod n. Receiver

1. For each received message Ri, calculate Ri s mod n. 2. Convert the string of digits back to a string of characters. Why does this method work? Well, we know that U(n) < U( p) % U(q) < Zp21 % Zq21. Thus an element of the form xm in U(n) corresponds under an isomorphism to one of the form (mx1, mx2) in Zp21 % Zq21. Since m is the least common multiple of p 2 1 and q 2 1, we may write m 5 u( p 2 1) and m 5 v(q 2 1) for some u and v. Then (mx1, mx2) 5 (u( p 2 1)x1, v(q 2 1)x2) 5 (0, 0) in Zp21 % Zq21, and it follows that xm 5 1 for all x in U(n). So, because each message Mi is an element of U(n) and r was chosen so that rs 5 1 1 tm for some t, we have, modulo n, Ri s 5 (Mi r ) s 5 Mi rs 5 Mi 11tm 5 (Mi m)tMi 5 1tMi 5 Mi. In 2002, Ronald Rivest, Adi Shamir, and Leonard Adleman received the Association for Computing Machinery A. M. Turing Award which is considered the “Nobel Prize of Computing” for their contribution to public key cryptography. The software for Computer Exercise 5 in this chapter implements the RSA scheme for small primes. Digital Signatures With so many financial transactions now taking place electronically, the problem of authenticity is paramount. How is a stockbroker to know that an electronic message she receives that tells her to sell one stock and buy another actually came from her client? The technique used in public key cryptography allows for digital signatures as well. Let us say that person A wants to send a secret message to person B in such a way that only B can decode the message and B will know that only A could have sent it. Abstractly, let EA and DA denote the algorithms that A uses for encryption and decryption, respectively, and let EB and DB denote the algorithms that B uses for encryption and decryption, respectively. Here we assume that EA and EB are available to the public, whereas DA is known only to A and DB is known only to B and that DBEB and EADA applied to any message leaves the message unchanged. Then A sends

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a message M to B as EB (DA(M)) and B decodes the received message by applying the function EADB to it to obtain (EADB) (EB (DA(M)) 5 EA(DBEB )(DA(M)) 5 EA(DA(M)) 5 M. Notice that only A can execute the first step [i.e., create DA(M)] and only B can implement the last step (i.e., apply EADB to the received message). Transactions using digital signatures became legally binding in the United States in October 2000. Application to Genetics† The genetic code can be conveniently modeled using elements of Z4 % Z4 % ? ? ? % Z4 where we omit the parentheses and the commas and just use strings of 0s, 1s, 2s, and 3s and add componentwise modulo 4. A DNA molecule is composed of two long strands in the form of a double helix. Each strand is made up of strings of the four nitrogen bases adenine (A), thymine (T), guanine (G), and cytosine (C). Each base on one strand binds to a complementary base on the other strand. Adenine always is bound to thymine, and guanine always is bound to cytosine. To model this process, we identify A with 0, T with 2, G with 1, and C with 3. Thus, the DNA segment ACGTAACAGGA and its complement segment TGCATTGTCCT are denoted by 03120030110 and 21302212332. Noting that in Z4, 0 1 2 5 2, 2 1 2 5 0, 1 1 2 5 3, and 3 1 2 5 1, we see that adding 2 to elements of Z4 interchanges 0 and 2 and 1 and 3. So, for any DNA segment a1a2 ? ? ? an represented by elements of Z4 % Z4 % ? ? ? % Z4, we see that its complementary segment is represented by a1a2 ? ? ? an 1 22 ? ? ? 2. Application to Electric Circuits Many homes have light fixtures that are operated by a pair of switches. They are wired so that when either switch is thrown the light changes its status (from on to off or vice versa). Suppose the wiring is done so that the light is on when both switches are in the up position. We can conveniently think of the states of the two switches as being matched with the elements of Z2 % Z2 with the two switches in the up position corresponding to (0, 0) and the two switches in the down position corresponding to (1, 1). Each time a switch is thrown, we add 1 to the corresponding component in the group Z2 % Z2. We then see that the lights are on when the switches correspond to the elements of the subgroup (1, 1) and are off when the switches correspond to the elements †This

discussion is adapted from [3].

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in the coset (1, 0) 1 (1, 1). A similar analysis applies in the case of three switches with the subgroup {(0, 0, 0), (1, 1, 0), (0, 1, 1), (1, 0, 1)} corresponding to the lights-on situation.

Exercises What’s the most difficult aspect of your life as a mathematician, Diane Maclagan, an assistant professor at Rutgers, was asked. “Trying to prove theorems,” she said. And the most fun? “Trying to prove theorems.”

1. Prove that the external direct product of any finite number of groups is a group. (This exercise is referred to in this chapter.) 2. Show that Z2 % Z2 % Z2 has seven subgroups of order 2. 3. Let G be a group with identity eG and let H be a group with identity eH. Prove that G is isomorphic to G % {eH} and that H is isomorphic to {eG} % H. 4. Show that G % H is Abelian if and only if G and H are Abelian. State the general case. 5. Prove or disprove that Z % Z is a cyclic group. 6. Prove, by comparing orders of elements, that Z8 % Z2 is not isomorphic to Z4 % Z4. 7. Prove that G1 % G2 is isomorphic to G2 % G1. State the general case. 8. Is Z3 % Z9 isomorphic to Z27? Why? 9. Is Z3 % Z5 isomorphic to Z15? Why? 10. How many elements of order 9 does Z3 % Z9 have? (Do not do this exercise by brute force.) 11. How many elements of order 4 does Z4 % Z4 have? (Do not do this by examining each element.) Explain why Z4 % Z4 has the same number of elements of order 4 as does Z8000000 % Z400000. Generalize to the case Z4m % Z4n. 12. The dihedral group Dn of order 2n (n $ 3) has a subgroup of n rotations and a subgroup of order 2. Explain why Dn cannot be isomorphic to the external direct product of two such groups. 13. Prove that the group of complex numbers under addition is isomorphic to R % R. 14. Suppose that G1 < G2 and H1 < H2. Prove that G1 % H1 < G2 % H2. State the general case. 15. If G % H is cyclic, prove that G and H are cyclic. State the general case.

168

Groups

16. In Z40 % Z30, find two subgroups of order 12. 17. If r is a divisor of m and s is a divisor of n, find a subgroup of Zm % Zn isomorphic to Zr % Zs. 18. Find a subgroup of Z12 % Z18 isomorphic to Z9 % Z4. 19. Let G and H be finite groups and (g, h) [ G % H. State a necessary and sufficient condition for (g, h) 5 g % h. 20. Determine the number of elements of order 15 and the number of cyclic subgroups of order 15 in Z30 % Z20. 21. What is the order of any nonidentity element of Z3 % Z3 % Z3? Generalize. 22. Let m . 2 be an even integer and let n . 2 be an odd integer. Find a formula for the number of elements of order 2 in Dm % Dn. 23. Let M be the group of all real 2 3 2 matrices under addition. Let N 5 R % R % R % R under componentwise addition. Prove that M and N are isomorphic. What is the corresponding theorem for the group of m 3 n matrices under addition? 24. The group S3 % Z2 is isomorphic to one of the following groups: Z12, Z6 % Z2, A4, D6. Determine which one by elimination. 25. Let G be a group, and let H 5 {(g, g) | g [ G}. Show that H is a subgroup of G % G. (This subgroup is called the diagonal of G % G.) When G is the set of real numbers under addition, describe G % G and H geometrically. 26. Find a subgroup of Z4 % Z2 that is not of the form H % K, where H is a subgroup of Z4 and K is a subgroup of Z2. 27. Find all subgroups of order 3 in Z9 % Z3. 28. Find all subgroups of order 4 in Z4 % Z4. 29. What is the largest order of any element in Z30 % Z20? 30. How many elements of order 2 are in Z2000000 % Z4000000? Generalize. 31. Find a subgroup of Z800 % Z200 that is isomorphic to Z2 % Z4. 32. Find a subgroup of Z12 % Z4 % Z15 that has order 9. 33. Prove that R* % R* is not isomorphic to C*. (Compare this with Exercise 13.) 34. Let 1 H 5 • £0 0

a 1 0

b 0 § † a, b 1

[ Z3 ¶ .

(See Exercise 36 in Chapter 2 for the definition of multiplication.) Show that H is an Abelian group of order 9. Is H isomorphic to Z9 or to Z3 % Z3?

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35. Let G 5 {3m6n | m, n [ Z} under multiplication. Prove that G is isomorphic to Z % Z. Does your proof remain valid if G 5 {3m9n | m, n [ Z}? 36. Let (a1, a2, . . . , an) [ G1 % G2 % ? ? ? % Gn. Give a necessary and sufficient condition for |(a1, a2, . . . , an)| 5 `. 37. Prove that D3 % D4 ] D12% Z2. 38. Determine the number of cyclic subgroups of order 15 in Z90 % Z36. 39. If a group has exactly 24 elements of order 6, how many cyclic subgroups of order 6 does it have? 40. For any Abelian group G and any positive integer n, let Gn 5 {gn | g [ G} (see Exercise 15, Supplementary Exercises for Chapters 1– 4). If H and K are Abelian, show that (H % K)n 5 Hn % Kn. 41. Express Aut(U(25)) in the form Zm % Zn. 42. Determine Aut(Z2 % Z2). 43. Suppose that n1, n2, . . . , nk are positive even integers. How many elements of order 2 does Zn % Zn % . . . % Zn have ? How many are 1 2 k there if we drop the requirement that n1, n2, . . . , nk must be even? 44. Is Z10 % Z12 % Z6 ^ Z60 % Z6 % Z2? 45. Is Z10 % Z12 % Z6 ^ Z15 % Z4 % Z12? 46. Find an isomorphism from Z12 to Z4 % Z3. 47. How many isomorphisms are there from Z12 to Z4 % Z3? 48. Suppose that f is an isomorphism from Z3 % Z5 to Z15 and f(2, 3) 5 2. Find the element in Z3 % Z5 that maps to 1. 49. Let (a, b) belong to Zm % Zn. Prove that |(a, b)| divides lcm(m, n). 50. Let G 5 {ax2 1 bx 1 c | a, b, c [ Z3}. Add elements of G as you would polynomials with integer coefficients, except use modulo 3 addition. Prove that G is isomorphic to Z3 % Z3 % Z3. Generalize. 51. Use properties of U-groups to determine all cyclic groups that have exactly two generators. 52. Explain a way that a string of length n of the four nitrogen bases A, T, G, and C could be modeled with the external direct product of n copies of Z2 % Z2. 53. Let p be a prime. Prove that Zp % Zp has exactly p 1 1 subgroups of order p. 54. Give an example of an infinite non-Abelian group that has exactly six elements of finite order. 55. Give an example to show that there exists a group with elements a and b such that |a| 5 `, |b| 5 ` and |ab| 5 2. 56. Express U(165) as an external direct product of cyclic groups of the form Zn. 57. Express U(165) as an external direct product of U-groups in four different ways.

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58. Without doing any calculations in Aut(Z20), determine how many elements of Aut(Z20) have order 4. How many have order 2? 59. Without doing any calculations in Aut(Z720), determine how many elements of Aut(Z720) have order 6. 60. Without doing any calculations in U(27), decide how many subgroups U(27) has. 61. What is the largest order of any element in U(900)? 62. Let p and q be odd primes and let m and n be positive integers. Explain why U( pm) % U(qn) is not cyclic. 63. Use the results presented in this chapter to prove that U(55) is isomorphic to U(75). 64. Use the results presented in this chapter to prove that U(144) is isomorphic to U(140). 65. For every n . 2, prove that U(n)2 5 {x2 | x [ U(n)} is a proper subgroup of U(n). 66. Show that U(55)3 5 {x3 | x [ U(55)} is U(55). 67. Find an integer n such that U(n) contains a subgroup isomorphic to Z5 % Z5. 68. Find a subgroup of order 6 in U(700). 69. Show that there is a U-group containing a subgroup isomorphic to Z3 % Z3. 70. Show that no U-group has order 14. 71. Show that there is a U-group containing a subgroup isomorphic to Z14. 72. Show that no U-group is isomorphic to Z4 % Z4. 73. Show that there is a U-group containing a subgroup isomorphic to Z4 % Z4. 74. Using the RSA scheme with p 5 37, q 5 73, and r 5 5, what number would be sent for the message “RM”? 75. Assuming that a message has been sent via the RSA scheme with p 5 37, q 5 73, and r 5 5, decode the received message “34.”

Computer Exercises The geek shall inherit the earth. LEV GROSSMAN

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian

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1. This software lists the elements of Us(st), where s and t are relatively prime. Run the program for (s, t) 5 (5, 16), (16, 5), (8, 25), (5, 9), (9, 5), (9, 10), (10, 9), and (10, 25). 2. This software computes the elements of the subgroup U(n)k 5 {xk | x [ U(n)} of U(n) and its order. Run the program for (n, k) 5 (27, 3), (27, 5), (27, 7), and (27, 11). Do you see a relationship connecting |U(n)| and |U(n)k|, f(n), and k? Make a conjecture. Run the program for (n, k) 5 (25, 3), (25, 5), (25, 7), and (25, 11). Do you see a relationship connecting |U(n)| and |U(n)k|, f(n), and k? Make a conjecture. Run the program for (n, k) 5 (32, 2), (32, 4), and (32, 8). Do you see a relationship connecting |U(n)| and |U(n)k|, f(n), and k? Make a conjecture. Is your conjecture valid for (32, 16)? If not, restrict your conjecture. Run the program for (n, k) 5 (77, 2), (77, 3), (77, 5), (77, 6), (77, 10), and (77, 15)? Do you see a relationship among U(77, 6), U(77, 2), and U(77, 3)? What about U(77, 10) U(77, 2), and U(77, 5)? What about U(77, 15), U(77, 3), and U(77, 5)? Make a conjecture. Use the theory developed in this chapter about expressing U(n) as external direct products of cyclic groups of the form Zn to analyze these groups to verify your conjectures. 3. This software implements the algorithm given on page 160 to express U(n) as an external direct product of groups of the form Zk. Run the program for n 5 3 ? 5 ? 7, 16 ? 9 ? 5, 8 ? 3 ? 25, 9 ? 5 ? 11, and 2 ? 27 ? 125. 4. This software allows you to input positive integers n1, n2, n3, . . . , nk, where k # 5, and compute the number of elements in Zn % Zn % ? ? ? % Zn of any specified order m. Use this software to verify the values obtained in Examples 4 and 5 and in Exercise 20. Run the software for n1 5 6, n2 5 10, n3 5 12, and m 5 6. 5. This program implements the RSA public key cryptography scheme. The user enters two primes p and q, an r that is relatively prime to m 5 lcm (p 2 1, q 2 1), and the message M to be sent. Then the program computes s, which is the inverse of r mod m, and the value of Mr mod pq. Also, the user can input those numbers and have the computer raise the numbers to the s power to obtain the original input. 6. This software determines the order of Aut(Zp % Zp), where p is a prime. Run the software for p 5 3, 5, and 7. Is the result always divisible by p? Is the result always divisible by p 2 1? Is the result always divisible by p 1 1? Make a conjecture about the order of Aut(Zp % Zp) for all primes p. 1

2

k

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Groups

7. This software determines the order of Aut(Zp % Zp % Zp), where p is a prime. Run the software for p 5 3, 5, and 7. What is the highest power of p that divides the order? What is the highest power of p 2 1 that divides the order? What is the highest power of p 1 1 that divides the order? Make a conjecture about the order of Aut(Zp % Zp % Zp) for all primes p.

References 1. J. A. Gallian and D. Rusin, “Factoring Groups of Integers Modulo n,” Mathematics Magazine 53 (1980): 33–36. 2. D. Shanks, Solved and Unsolved Problems in Number Theory, 2nd ed., New York: Chelsea, 1978. 3. S. Washburn, T. Marlowe, and C. Ryan, Discrete Mathematics, Reading, MA: Addison-Wesley, 1999.

Suggested Readings Y. Cheng, “Decomposition of U-groups,” Mathematics Magazine 62 (1989): 271–273. This article explores the decomposition of U(st), where s and t are relatively prime, in greater detail than we have provided. David J. Devries, “The Group of Units in Zn,” Mathematics Magazine 62 (1989): 340. This article provides a simple proof that U(n) is not cyclic when n is not of the form 1, 2, 4, pk, or 2pk, where p is an odd prime. David R. Guichard, “When Is U(n) Cyclic? An Algebra Approach,” Mathematics Magazine 72 (1999): 139–142. The author provides a group-theoretic proof of the fact that U(n) is cyclic if and only if n is 1, 2, 4, pk, or 2pk, where p is an odd prime.

Leonard Adleman “. . . Dr. Adleman [has played] a central role in some of the most surprising, and provocative, discoveries in theoretical computer science.” GINA KOLATA, The New York Times,

13 December 1994.

LEONARD ADLEMAN grew up in San Francisco. He did not have any great ambitions for himself and, in fact, never even thought about becoming a mathematician. He enrolled at the University of California at Berkeley intending to be a chemist, then changed his mind and said he would be a doctor. Finally, he settled on a mathematics major. “I had gone through a zillion things and finally the only thing that was left where I could get out in a reasonable time was mathematics,” he said. Adleman graduated in five years, in 1968, “wondering what I wanted to do with my life.” He took a job as a computer programmer at the Bank of America. Then he decided that maybe he should be a physicist, so he began taking classes at San Francisco State College while working at the bank. Once again, Adleman lost interest. “I didn’t like doing experiments, I liked thinking about things,” he said. Later, he returned to Berkeley with the aim of getting a Ph.D. in computer science. “I thought that getting a Ph.D. in computer science would at least further my career,” he said. But, while in graduate school, something else happened to Adleman. He finally understood the true nature and compelling beauty of mathematics. He discovered, he said, that mathematics “is less related to accounting than it is to philosophy.”

“People think of mathematics as some kind of practical art,” Adleman said. But, he added, “the point when you become a mathematician is where you somehow see through this and see the beauty and power of mathematics.” Adleman got his Ph.D. in 1976 and immediately landed a job as an assistant professor of mathematics at the Massachusetts Institute of Technology. There he met Ronald Rivest and Adi Shamir, who were trying to invent an unbreakable public key system. They shared their excitement about the idea with Adleman, who greeted it with a polite yawn, thinking it impractical and not very interesting. Nevertheless, Adleman agreed to try to break the codes Rivest and Shamir proposed. Rivest and Shamir invented 42 coding systems, and each time Adleman broke the code. Finally, on their 43rd attempt, they hit upon what is now called the RSA scheme. Adleman’s mode of working is to find something that intrigues him and to dig in. He does not read mathematics journals, he says, because he does not want to be influenced by other people’s ideas. Asked what it is like to simply sit and think for six months, Adleman responded, “That’s what a mathematician always does. Mathematicians are trained and inclined to sit and think. A mathematician can sit and think intensely about a problem for 12 hours 173

a day, six months straight, with perhaps just a pencil and paper.” The only prop he needs, he said, is a blackboard to stare at. Adapted from an article by Gina Kolata, The New York Times, 13 December 1994.

For more information about Adleman, visit: http://www.wikipedia.com

Supplementary Exercises for Chapters 5–8 My mind rebels at stagnation. Give me problems, give me work, give me the most obstruse cryptogram, or the most intricate analysis, and I am in my own proper atmosphere. SHERLOCK HOLMES, The Sign of Four

True/False questions for Chapters 5–8 are available on the Web at: www.d.umn.edu/~jgallian/TF 1. A subgroup N of a group G is called a characteristic subgroup if f(N) 5 N for all automorphisms f of G. (The term characteristic was first applied by G. Frobenius in 1895.) Prove that every subgroup of a cyclic group is characteristic. 2. Prove that the center of a group is characteristic. 3. The commutator subgroup G9 of a group G is the subgroup generated by the set {x21y21xy | x, y [ G}. (That is, every element of G9 has the form a1i1a2i2 ? ? ? a kik, where each aj has the form x21y21xy, each ij 5 61, and k is any positive integer.) Prove that G9 is a characteristic subgroup of G. (This subgroup was first introduced by G. A. Miller in 1898.) 4. Prove that the property of being a characteristic subgroup is transitive. That is, if N is a characteristic subgroup of K and K is a characteristic subgroup of G, then N is a characteristic subgroup of G. 5. Let G 5 Z3 % Z3 % Z3 and let H be the subgroup of SL(3, Z3) consisting of 1 H 5 • £0 0

174

a 1 0

b 0 § † a, b c [ Z3 ¶ . 1

8 | Supplementary Exercises for Chapters 5–8

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(See Exercise 36 in Chapter 2 for the definition of multiplication.) Determine the number of elements of each order in G and H. Are G and H isomorphic? (This exercise shows that two groups with the same number of elements of each order need not be isomorphic.) 6. Let H and K be subgroups of a group G and let HK 5 {hk | h [ H, k [ K} and KH 5 {kh | k [ K, h [ H}. Prove that HK is a group if and only if HK 5 KH. 7. Let H and K be subgroups of a finite group G. Prove that 0 HK 0 5 8.

9. 10. 11.

12. 13. 14. 15. 16. 17. 18.

19. 20.

0H0 0K0 . 0H d K0

(This exercise is referred to in Chapters 10, 11, and 24.) The exponent of a group is the smallest positive integer n such that xn 5 e for all x in the group. Prove that every finite group has an exponent that divides the order of the group. Determine all U-groups of exponent 2. Suppose that H and K are subgroups of a group and that |H| and |K| are relatively prime. Show that H y K 5 {e}. Let R1 denote the multiplicative group of positive real numbers and let T 5 {a 1 bi [ C*| a2 1 b2 5 1} be the multiplicative group of complex numbers of norm 1. Show that every element of C* can be uniquely expressed in the form rz, where r [ R1 and z [ T. Use a group-theoretic proof to show that Q* under multiplication is not isomorphic to R* under multiplication. Use a group-theoretic proof to show that Q under addition is not isomorphic to R under addition. Prove that R under addition is not isomorphic to R* under multiplication. Show that Q1 (the set of positive rational numbers) under multiplication is not isomorphic to Q under addition. Suppose that G 5 {e, x, x2, y, yx, yx2} is a non-Abelian group with |x| 5 3 and |y| 5 2. Show that xy 5 yx2. Let p be an odd prime. Show that 1 is the only solution of x p22 5 1 in U(p). Let G be an Abelian group under addition. Let n be a fixed positive integer and let H 5 {(g, ng) | g [ G}. Show that H is a subgroup of G % G. When G is the set of real numbers under addition, describe H geometrically. Find five subgroups of Z12 % Z20 1 Z10 isomorphic to Z4 % Z5. Suppose that G 5 G1 % G2 % ? ? ? % Gn. Prove that Z(G) 5 Z(G1) % Z(G2) % ? ? ? % Z(Gn).

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Groups

21. Exhibit four nonisomorphic groups of order 18. 22. What is the order of the largest cyclic subgroup in Aut(Z720)? (Hint: It is not necessary to consider automorphisms of Z720.) 23. Let G be the group of all permutations of the positive integers. Let H be the subset of elements of G that can be expressed as a product of a finite number of cycles. Prove that H is a subgroup of G. 24. Let H be a subgroup of G. Show that Z(G)H is a subgroup of G. 25. Show that D11 % Z3 ] D3 % Z11. (This exercise is referred to in Chapter 24.) 26. Show that D33 ] D11 % Z3. (This exercise is referred to in Chapter 24.) 27. Show that D33 ] D3 % Z11. (This exercise is referred to in Chapter 24.) 28. Exhibit four nonisomorphic groups of order 66. (This exercise is referred to in Chapter 24.) 29. Prove that |Inn(G)| 5 1 if and only if G is Abelian. 30. Prove that x100 5 1 for all x in U(1000). 31. Find a subgroup of order 6 in U(450). 32. List four elements of Z20 % Z5 % Z60 that form a noncyclic subgroup. 33. In S10, let b 5 (13)(17)(265)(289). Find an element in S10 that commutes with b but is not a power of b. 34. Prove or disprove that Z4 % Z15 < Z6 % Z10. 35. Prove or disprove that D12 < Z3 % D4. 36. Describe a three-dimensional solid whose symmetry group is isomorphic to D5. 37. Let G 5 U(15) % Z10 % S5. Find the order of (2, 3, (123)(15)). Find the inverse of (2, 3, (123)(15)). 38. Let G 5 Z % Z10 and let H 5 {g [ G| |g| 5 ` or |g| 5 1}. Prove or disprove that H is a subgroup of G. 39. Let G be an infinite group of the form G1 % G2 % . . . % Gn where each Gi is a nontrivial group and n . 1. Prove that G is not cyclic. 40. For any s in Sn and any k-cycle (i1i2 . . . ik) in Sn, prove that s(i1i2 . . . ik)s21 5 s(i1)s(i2) . . . s(ik). 41. Find an element of order 10 in A9. 42. In the left regular representation for D4, write TR90 and TH in matrix form and in cycle form. 43. How many elements of order 6 are in S7? 44. Prove that S3 % S4 is not isomorphic to a subgroup of S6. 45. Find a permutation b such that b2 5 (13579)(268).

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46. In R % R under componentwise addition, let H 5 {(x, 3x) | x [ R}. (Note that H is the subgroup of all points on the line y 5 3x.) Show that (2, 5) 1 H is a straight line passing through the point (2, 5) and parallel to the line y 5 3x. 47. In R % R, suppose that H is the subgroup of all points lying on a line through the origin. Show that any left coset of H is either H or a line parallel to H. 48. Let G be a group of permutations on the set {1, 2, . . . , n}. Recall that stabG(1) 5 {a [ G | a(1) 5 1}. If g sends 1 to k, prove that g stabG(1) 5 {b [ G | b(1) 5 k}. 49. Let H be a subgroup of G and let a, b [ G. Show that aH 5 bH if and only if Ha21 5 Hb21. 50. Suppose that G is a finite Abelian group that does not contain a subgroup isomorphic to Zp % Zp for any prime p. Prove that G is cyclic. 51. Let p be a prime. Determine the number of elements of order p in Zp2 % Zp2. 52. Show that Zp2 % Zp2 has exactly one subgroup isomorphic to Zp % Zp. 53. Let p be a prime. Determine the number of subgroups of Zp2 % Zp2 isomorphic to Zp2. 54. Find a group of order 32 ? 52 ? 72 ? 28 that contains a subgroup isomorphic to A8. 55. Let p and q be distinct odd primes. Let n 5 lcm(p 2 1, q 2 1). Prove that xn 5 1 for all x [ U( pq). 56. Prove that D6 is not isomorphic to a subgroup of S4. 57. Prove that the permutations (12) and (123 . . . n) generate Sn. (That is, every member of Sn can be expressed as some combination of these elements. 58. Suppose that n is even and s is an (n 2 1)-cycle in Sn. Show that s does not commute with any element of order 2. 59. Suppose that n is odd and s is an n-cycle in Sn. Prove that s does not commute with any element of order 2.

Normal Subgroups 9 and Factor Groups It is tribute to the genius of Galois that he recognized that those subgroups for which the left and right cosets coincide are distinguished ones. Very often in mathematics the crucial problem is to recognize and to discover what are the relevant concepts; once this is accomplished the job may be more than half done. I. N. HERSTEIN, Topics in Algebra

Normal Subgroups As we saw in Chapter 7, if G is a group and H is a subgroup of G, it is not always true that aH 5 Ha for all a in G. There are certain situations where this does hold, however, and these cases turn out to be of critical importance in the theory of groups. It was Galois, about 175 years ago, who first recognized that such subgroups were worthy of special attention. Definition Normal Subgroup

A subgroup H of a group G is called a normal subgroup of G if aH 5 Ha for all a in G. We denote this by H v G.

Many students make the mistake of thinking that “H is normal in G” means ah 5 ha for a [ G and h [ H. This is not what normality of H means; rather, it means that if a [ G and h [ H, then there exist elements h9 and h0 in H such that ah 5 h9a and ha 5 ah0. Think of it this way: You can switch the order of a product of an element from the group and an element from the normal subgroup, but you must “fudge” a bit on the element from the normal subgroup by using h9 or h0 rather than h. (It is possible that h9 5 h or h0 5 h, but we may not assume this.) There are several equivalent formulations of the definition of normality. We have chosen the one that is the easiest to use in applications. However, to verify that a subgroup is normal, it is usually better to use Theorem 9.1, which is a weaker version of property 7 of the lemma in Chapter 7. It allows us to substitute a condition about two subgroups of G for a condition about two cosets of G.

178

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179

Theorem 9.1 Normal Subgroup Test A subgroup H of G is normal in G if and only if xHx21 # H for all x in G.

PROOF If H is normal in G, then for any x [ G and h [ H there is an h9 in H such that xh 5 h9x. Thus, xhx21 5 h9, and therefore xHx21 # H. Conversely, if xHx21 # H for all x, then, letting x 5 a, we have aHa21 # H or aH # Ha. On the other hand, letting x 5 a21, we have a21H(a21)21 5 a21Ha # H or Ha # aH. EXAMPLE 1 Every subgroup of an Abelian group is normal. (In this case, ah 5 ha for a in the group and h in the subgroup.) EXAMPLE 2 The center Z(G) of a group is always normal. [Again, ah 5 ha for any a [ G and any h [ Z(G).] EXAMPLE 3 The alternating group An of even permutations is a normal subgroup of Sn. [Note, for example, that for (12) [ Sn and (123) [ An, we have (12)(123) 2 (123)(12) but (12)(123) 5 (132)(12) and (132) [ An.] EXAMPLE 4 The subgroup of rotations in Dn is normal in Dn. (For any rotation r and any reflection f, we have fr 5 r21f, whereas for any rotations r and r9, we have rr9 5 r9r.) EXAMPLE 5 The group SL(2, R) of 2 3 2 matrices with determinant 1 is a normal subgroup of GL(2, R), the group of 2 3 2 matrices with nonzero determinant. To verify this, we use the normal subgroup test given in Theorem 9.1. Let x [ GL(2, R) 5 G, h [ SL(2, R) 5 H and note that det xhx21 5 (det x)(det h)(det x)21 5 (det x)(det x)21 5 1. So, xhx21 [ H, and, therefore, xHx21 # H. EXAMPLE 6 Referring to the group table for A4 given in Table 5.1 on page 107, we may observe that H 5 {a1, a2, a3, a4} is a normal subgroup of A4 , whereas K 5 {a1, a5, a9} is not a normal subgroup of A4. To see that H is normal, simply note that for any b in A4, bHb21 is a subgroup of order 4 and H is the only subgroup of A4 of order 4 since all other elements of A4 have order 3. Thus, bHb21 5 H. In con21 s K. trast, a2a5a21 2 5 a7, so that a2Ka 2

180

Groups

Factor Groups We have yet to explain why normal subgroups are of special significance. The reason is simple. When the subgroup H of G is normal, then the set of left (or right) cosets of H in G is itself a group—called the factor group of G by H (or the quotient group of G by H). Quite often, one can obtain information about a group by studying one of its factor groups. This method will be illustrated in the next section of this chapter. Theorem 9.2 Factor Groups (O. Hölder, 1889) Let G be a group and let H be a normal subgroup of G. The set G/H 5 {aH | a [ G} is a group under the operation (aH)(bH) 5 abH.†

PROOF Our first task is to show that the operation is well defined; that is, we must show that the correspondence defined above from G/H 3 G/H into G/H is actually a function. To do this we assume that for some elements a, a9, b, and b9 from G, we have aH 5 a9H and bH 5 b9H and verify that aHbH 5 a9Hb9H. That is, verify that abh 5 a9b9H. (This shows that the definition of multiplication depends only on the cosets and not on the coset representatives.) From aH 5 a9H and bH 5 b9H , we have a9 5 ah1 and b9 5 bh2 for some h1, h2 in H, and therefore a9b9H 5 ah1bh2H 5 ah1bH 5 ah1Hb 5 aHb 5 abH. Here we have made multiple use of associativity, property 2 of the lemma in Chapter 7, and the fact that H v G. The rest is easy: eH 5 H is the identity; a21H is the inverse of aH; and (aHbH)cH 5 (ab)HcH 5 (ab)cH 5 a(bc)H 5 aH(bc)H 5 aH(bHcH). This proves that G/H is a group. Although it is merely a curiosity, we point out that the converse of Theorem 9.2 is also true; that is, if the correspondence aHbH 5 abH defines a group operation on the set of left cosets of H in G, then H is normal in G. The next few examples illustrate the factor group concept. EXAMPLE 7 Let 4Z 5 {0, 64, 68, . . .}. To construct Z/4Z, we first must determine the left cosets of 4Z in Z. Consider the following four cosets: 0 1 4Z 5 4Z 5 {0, 64, 68, . . .}, 1 1 4Z 5 {1, 5, 9, . . . ; 23, 27, 211, . . .},

†The

notation G/H was first used by C. Jordan.

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9 | Normal Subgroups and Factor Groups

2 1 4Z 5 {2, 6, 10, . . . ; 22, 26, 210, . . .}, 3 1 4Z 5 {3, 7, 11, . . . ; 21, 25, 29, . . .}. We claim that there are no others. For if k [ Z, then k 5 4q 1 r, where 0 # r , 4; and, therefore, k 1 4Z 5 r 1 4q 1 4Z 5 r 1 4Z. Now that we know the elements of the factor group, our next job is to determine the structure of Z/4Z. Its Cayley table is

0 1 4Z 1 1 4Z 2 1 4Z 3 1 4Z

0 1 4Z

1 1 4Z

2 1 4Z

3 1 4Z

0 1 4Z 1 1 4Z 2 1 4Z 3 1 4Z

1 1 4Z 2 1 4Z 3 1 4Z 0 1 4Z

2 1 4Z 3 1 4Z 0 1 4Z 1 1 4Z

3 1 4Z 0 1 4Z 1 1 4Z 2 1 4Z

Clearly, then, Z/4Z L Z4. More generally, if for any n . 0 we let nZ 5 {0, 6n, 62n, 63n, . . .}, then Z/nZ is isomorphic to Zn. EXAMPLE 8 Let G 5 Z18 and let H 5 6 5 {0, 6, 12}. Then G/H 5 {0 1 H, 1 1 H, 2 1 H, 3 1 H, 4 1 H, 5 1 H}. To illustrate how the group elements are combined, consider (5 1 H) 1 (4 1 H). This should be one of the six elements listed in the set G/H. Well, (5 1 H) 1 (4 1 H) 5 5 1 4 1 H 5 9 1 H 5 3 1 6 1 H 5 3 1 H, since H absorbs all multiples of 6. A few words of caution about notation are warranted here. When H is a normal subgroup of G, the expression |aH| has two possible interpretations. One could be thinking of aH as a set of elements and |aH| as the size of the set; or, as is more often the case, one could be thinking of aH as a group element of the factor group G/H and |aH| as the order of the element aH in G/H. In Example 8, for instance, the set 3 1 H has size 3, since 3 1 H 5 {3, 9, 15}. But the group element 3 1 H has order 2, since (3 1 H) 1 (3 1 H) 5 6 1 H 5 0 1 H. As is usually the case when one notation has more than one meaning, the appropriate interpretation will be clear from the context. EXAMPLE 9 Let _ 5 {R0, R180}, and consider the factor group of the dihedral group D4 (see page 31 for the multiplication table for D4) D4/_ 5 {_, R90_, H_, D_}. The multiplication table for D4/_ is given in Table 9.1. (Notice that even though R90H 5 D9, we have used D_ in Table 9.1 for H_R90_ because D9_ 5 D_.)

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Groups

Table 9.1

_ R90_ _ H_ _ D_

_

R90_

_ H_

_ D_

_ R90_ H_ D_

R90_ _ D_ H_

H_ D_ _ R90_

D_ H_ R90_ _

D4/_ provides a good opportunity to demonstrate how a factor group of G is related to G itself. Suppose we arrange the heading of the Cayley table for D4 in such a way that elements from the same coset of _ are in adjacent columns (Table 9.2). Then, the multiplication table for D4 can be blocked off into boxes that are cosets of _, and the substitution that replaces a box containing the element x with the coset x_ yields the Cayley table for D4/_. For example, when we pass from D4 to D4/_, the box H V

V H

in Table 9.2 becomes the element H_ in Table 9.1. Similarly, the box D D9

D9 D

becomes the element D_, and so on. Table 9.2 R0

R180

R90

R270

H

V

D

D9

R0 R180

R0 R180

R180 R0

R90 R270

R270 R90

H V

V H

D D9

D9 D

R90 R270

R90 R270

R270 R90

R180 R0

R0 R180

D9 D

D D9

H V

V H

H V

H V

V H

D D9

D9 D

R0 R180

R180 R0

R90 R270

R270 R90

D D9

D D9

D9 D

V H

H V

R270 R90

R90 R270

R0 R180

R180 R0

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9 | Normal Subgroups and Factor Groups

In this way, one can see that the formation of a factor group G/H causes a systematic collapse of the elements of G. In particular, all the elements in the coset of H containing a collapse to the single group element aH in G/H. EXAMPLE 10 Consider the group A4 as represented by Table 5.1 on page 107. (Here i denotes the permutation ai.) Let H 5 {1, 2, 3, 4}. Then the three cosets of H are H, 5H 5 {5, 6, 7, 8}, and 9H 5 {9, 10, 11, 12}. (In this case, rearrangement of the headings is unnecessary.) Blocking off the table for A4 into boxes that are cosets of H and replacing the boxes containing 1, 5, and 9 (see Table 9.3) with the cosets 1H, 5H, and 9H, we obtain the Cayley table for G/H given in Table 9.4. Table 9.3 1

2

3

4

5

6

7

8

9

10

11

12

1 2 3 4

1 2 3 4

2 1 4 3

3 4 1 2

4 3 2 1

5 6 7 8

6 5 8 7

7 8 5 6

8 7 6 5

9 10 11 12

10 9 12 11

11 12 9 10

12 11 10 9

5 6 7 8

5 6 7 8

8 7 6 5

6 5 8 7

7 8 5 6

9 10 11 12

12 11 10 9

10 9 12 11

11 12 9 10

1 2 3 4

4 3 2 1

2 1 4 3

3 4 1 2

9 10 11 12

9 10 11 12

11 12 9 10

12 11 10 9

10 9 12 11

1 2 3 4

3 4 1 2

4 3 2 1

2 1 4 3

5 6 7 8

7 8 5 6

8 7 6 5

6 5 8 7

Table 9.4

1H 5H 9H

1H

5H

9H

1H 5H 9H

5H 9H 1H

9H 1H 5H

This procedure can be illustrated more vividly with colors. Let’s say we had printed the elements of H in green, the elements of 5H in red, and the elements of 9H in blue. Then, in Table 9.3, each box would consist of elements of a uniform color. We could then think of

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Groups

the factor group as consisting of the three colors that define a group table isomorphic to G/H.

Green Red Blue

Green

Red

Blue

Green Red Blue

Red Blue Green

Blue Green Red

It is instructive to see what happens if we attempt the same procedure with a group G and a subgroup H that is not normal in G—that is, if we arrange the headings of the Cayley table so that the elements from the same coset of H are in adjacent columns and attempt to block off the table into boxes that are also cosets of H to produce a Cayley table for the set of cosets. Say, for instance, we were to take G to be A4 and H 5 {1, 5, 9}. The cosets of H would be H, 2H 5 {2, 6, 10}, 3H 5 {3, 7, 11}, and 4H 5 {4, 8, 12}. Then the first three rows of the rearranged Cayley table for A4 would be

1 5 9

1

5

9

2

6

10

3

7

11

4

8

12

1 5 9

5 9 1

9 1 5

2 8 11

6 12 3

10 4 7

3 6 12

7 10 4

11 2 8

4 7 10

8 11 2

12 3 6

But already we are in trouble, for blocking these off into 3 3 3 boxes yields boxes that contain elements of different cosets. Hence, it is impossible to represent an entire box by a single element of the box in the same way we could for boxes made from the cosets of a normal subgroup. Had we printed the rearranged table in four colors with all members of the same coset having the same color, we would see multicolored boxes rather than the uniformly colored boxes produced by a normal subgroup. In Chapter 11, we will prove that every finite Abelian group is isomorphic to a direct product of cyclic groups. In particular, an Abelian group of order 8 is isomorphic to one of Z8, Z4 % Z2, or Z2 % Z2 % Z2. In the next two examples, we examine Abelian factor groups of order 8 and determine the isomorphism type of each. EXAMPLE 11 Let G 5 U(32) 5 {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31} and H 5 U16(32) 5 {1, 17}. Then G/H is an Abelian group of order 16/2 5 8. Which of the three Abelian groups of order 8 is it—Z8, Z4 % Z2, or Z2 % Z2 % Z2? To answer this question, we need

9 | Normal Subgroups and Factor Groups

185

only determine the elements of G/H and their orders. Observe that the eight cosets 1H 5 {1, 17}, 3H 5 {3, 19}, 5H 5 {5, 21}, 9H 5 {9, 25}, 11H 5 {11, 27}, 13H 5 {13, 29},

7H 5 {7, 23}, 15H 5 {15, 31}

are all distinct, so that they form the factor group G/H. Clearly, (3H)2 5 9H 2 H, and so 3H has order at least 4. Thus, G/H is not Z2 % Z2 % Z2. On the other hand, direct computations show that both 7H and 9H have order 2, so that G/H cannot be Z8 either, since a cyclic group of even order has exactly one element of order 2 (Theorem 4.4). This proves that U(32)/U16(32) L Z4 % Z2, which (not so incidentally!) is isomorphic to U(16). EXAMPLE 12 Let G 5 U(32) and K 5 {1, 15}. Then |G/K| 5 8, and we ask, which of the three Abelian groups of order 8 is G/K? Since (3K)4 5 81K 5 17K 2 K, |3K| 5 8. Thus, G/K L Z8. It is crucial to understand that when we factor out by a normal subgroup H, what we are essentially doing is defining every element in H to be the identity. Thus, in Example 9, we are making R180_ 5 _ the identity. Likewise, R270_ 5 R90R180_ 5 R90_. Similarly, in Example 7, we are declaring any multiple of 4 to be 0 in the factor group Z/4Z. This is why 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z, and so on. In Example 11, we have 3H 5 19H, since 19 5 3 ? 17 in U(32) and going to the factor group makes 17 the identity. Algebraists often refer to the process of creating the factor group G/H as “killing” H.

Applications of Factor Groups Why are factor groups important? Well, when G is finite and H 2 {e}, G/H is smaller than G, and its structure is usually less complicated than that of G. At the same time, G/H simulates G in many ways. In fact, we may think of a factor group of G as a less complicated approximation of G (similar to using the rational number 3.14 for the irrational number p). What makes factor groups important is that one can often deduce properties of G by examining the less complicated group G/H instead. We illustrate this by giving another proof that A4 has no subgroup of order 6. EXAMPLE 13 A4 Has No Subgroup of Order 6 The group A4 of even permutations on the set {1, 2, 3, 4} has no subgroup H of order 6. To see this, suppose that A4 does have a subgroup H

186

Groups

of order 6. By Exercise 7 in this chapter, we know that H v A4. Thus, the factor group A4/H exists and has order 2. Since the order of an element divides the order of the group, we have for all a [ A4 that a2 H 5 (aH)2 5 H. Thus, a2 [ H for all a in A4. Referring to the main diagonal of the group table for A4 given in Table 5.1 on page 107, however, we observe that A4 has nine different elements of the form a2, all of which must belong to H, a subgroup of order 6. This is clearly impossible, so a subgroup of order 6 cannot exist in A4.† The next three theorems illustrate how knowledge of a factor group of G reveals information about G itself. Theorem 9.3 The G/Z Theorem Let G be a group and let Z(G) be the center of G. If G/Z(G) is cyclic, then G is Abelian.

PROOF Let gZ(G) be a generator of the factor group G/Z(G), and let a, b [ G. Then there exist integers i and j such that aZ(G) 5 (gZ(G))i 5 giZ(G) and bZ(G) 5 (gZ(G)) j 5 g jZ(G). Thus, a 5 gix for some x in Z(G) and b 5 gjy for some y in Z(G). It follows then that ab 5 (gix)(g jy) 5 gi(xg j)y 5 gi(g jx)y 5 (gig j)(xy) 5 (g jgi)(yx) 5 (g jy)(gix) 5 ba. A few remarks about Theorem 9.3 are in order. First, our proof shows that a better result is possible: If G/H is cyclic, where H is a subgroup of Z(G), then G is Abelian. Second, in practice, it is the contrapositive of the theorem that is most often used—that is, if G is non-Abelian, then G/Z(G) is not cyclic. For example, it follows immediately from this statement and Lagrange’s Theorem that a non-Abelian group of order pq, where p and q are primes, must have a trivial center. Third, if G/Z(G) is cyclic, it must be trivial.

†How

often have I said to you that when you have eliminated the impossible, whatever remains, however improbable, must be the truth. Sherlock Holmes, The Sign of Four

9 | Normal Subgroups and Factor Groups

187

Theorem 9.4 G/Z(G) < Inn(G) For any group G, G/Z(G) is isomorphic to Inn(G).

PROOF Consider the correspondence from G/Z(G) to Inn(G) given by T : gZ(G) → fg [where, recall, fg(x) 5 gxg21 for all x in G]. First, we show that T is well defined. To do this, we assume that gZ(G) 5 hZ(G) and verify that fg 5 fh. (This shows that the image of a coset of Z(G) depends only on the coset itself and not on the element representing the coset.) From gZ(G) 5 hZ(G) , we have that h21g belongs to Z(G). Then, for all x in G, h21gx 5 xh21g. Thus, gxg21 5 hxh21 for all x in G, and, therefore, fg 5 fh. Reversing this argument shows that T is one-to-one, as well. Clearly, T is onto. That T is operation-preserving follows directly from the fact that fgfh 5 fgh for all g and h in G. As an application of Theorems 9.3 and 9.4, we may easily determine Inn(D6) without looking at Inn(D6)! EXAMPLE 14 We know from Example 11 in Chapter 3 that |Z(D6)| 5 2. Thus, |D6 /Z (D6)| 5 6. So, by our classification of groups of order 6 (Theorem 7.2), we know that Inn(D6) is isomorphic to D3 or Z6. Now, if Inn(D6) were cyclic, then, by Theorem 9.4, D6 /Z(D6) would be also. But then, Theorem 9.3 would tell us that D6 is Abelian. So, Inn(D6) is isomorphic to D3. The next theorem demonstrates one of the most powerful proof techniques available in the theory of finite groups—the combined use of factor groups and induction. Theorem 9.5 Cauchy’s Theorem for Abelian Groups Let G be a finite Abelian group and let p be a prime that divides the order of G. Then G has an element of order p.

PROOF Clearly, this statement is true for the case in which G has order 2. We prove the theorem by using the Second Principle of Mathematical Induction on |G|. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if |x| 5 m and m 5 qn, where q is prime, then |xn| 5 q. So let x be an element of G of some prime order q, say. If

188

Groups

q 5 p, we are finished; so assume that q 2 p. Since every subgroup of an Abelian group is normal, we may construct the factor group G 5 G/x. Then G is Abelian and p divides |G|, since |G| 5 |G|/q. By induction, then, G has an element—call it yx—of order p. The conclusion now follows from Exercise 65.

Internal Direct Products As we have seen, the external direct product provides a way of putting groups together into a larger group. It would be quite useful to be able to reverse this process—that is, to be able to start with a large group and break it down into a product of smaller groups. It is occasionally possible to do this. To this end, suppose that H and K are subgroups of some group G. We define the set HK 5 {hk | h [ H, k [ K}. EXAMPLE 15 In U(24) 5 {1, 5, 7, 11, 13, 17, 19, 23}, let H 5 {1, 17} and K 5 {1, 13}. Then, HK 5 {1, 13, 17, 5}, since 5 5 17 ? 13 mod 24. EXAMPLE 16 In S3, let H 5 {(1), (12)} and K 5 {(1), (13)}. Then, HK 5 {(1), (13), (12), (12)(13)} 5 {(1), (13), (12), (132)}. The student should be careful not to assume that the set HK is a subgroup of G; in Example 15 it is, but in Example 16 it is not. Definition Internal Direct Product of H and K

We say that G is the internal direct product of H and K and write G 5 H 3 K if H and K are normal subgroups of G and G 5 HK

and H > K 5 {e}.

The wording of the phrase “internal direct product” is easy to justify. We want to call G the internal direct product of H and K if H and K are subgroups of G, and if G is naturally isomorphic to the external direct product of H and K. One forms the internal direct product by starting with a group G and then proceeding to find two subgroups H and K within G such that G is isomorphic to the external direct product of H and K. (The definition ensures that this is the case—see Theorem 9.6.) On the other hand, one forms an external direct product by starting with any two groups H and K, related or not, and proceeding to produce the larger group H % K. The difference between the two products is that the internal direct product can be formed within G itself, using subgroups of G and the operation of G, whereas the external direct product can be formed with totally unrelated groups by creating a new set and a new operation. (See Figures 9.1 and 9.2.)

9 | Normal Subgroups and Factor Groups

189

G

H

e

K

Figure 9.1 For the internal direct product, H and K must be subgroups of the same group.

H

K

Figure 9.2 For the external direct product, H and K can be any groups.

Perhaps the following analogy with integers will be useful in clarifying the distinction between the two products of groups discussed in the preceding paragraph. Just as we may take any (finite) collection of integers and form their product, we may also take any collection of groups and form their external direct product. Conversely, just as we may start with a particular integer and express it as a product of certain of its divisors, we may be able to start with a particular group and factor it as an internal direct product of certain of its subgroups. EXAMPLE 17 In D6, the dihedral group of order 12, let F denote some reflection and let Rk denote a rotation of k degrees. Then, D6 5 {R0, R120, R240, F, R120F, R240F} 3 {R0, R180}. Students should be cautioned about the necessity of having all conditions of the definition of internal direct product satisfied to ensure that HK L H % K. For example, if we take G 5 S3,

H 5 (123),

and

K 5 (12),

then G 5 HK, and H > K 5 {(1)}. But G is not isomorphic to H % K, since, by Theorem 8.2, H % K is cyclic, whereas S3 is not. Note that K is not normal. A group G can also be the internal direct product of a collection of subgroups.

190

Groups

Definition Internal Direct Product H1 3 H2 3 ? ? ? 3 Hn

Let H1, H2, . . . , Hn be a finite collection of normal subgroups of G. We say that G is the internal direct product of H1, H2, . . . , Hn and write G 5 H1 3 H2 3 ? ? ? 3 Hn, if 1. G 5 H1H2 ? ? ? Hn 5 {h1h2 ? ? ? hn | hi [ Hi} 2. (H1H2 ? ? ? Hi) > Hi11 5 {e} for i 5 1, 2, . . . , n 2 1.

This definition is somewhat more complicated than the one given for two subgroups. The student may wonder about the motivation for it— that is, why should we want the subgroups to be normal and why is it desirable for each subgroup to be disjoint from the product of all previous ones? The reason is quite simple. We want the internal direct product to be isomorphic to the external direct product. As the next theorem shows, the conditions in the definition of internal direct product were chosen to ensure that the two products are isomorphic. Theorem 9.6 H1 3 H2 3 ? ? ? 3 Hn L H1 % H2 % ? ? ? % Hn If a group G is the internal direct product of a finite number of subgroups H1, H2, . . . , Hn, then G is isomorphic to the external direct product of H1, H2, . . . , Hn.

PROOF We first show that the normality of the H’s together with the second condition of the definition guarantees that h’s from different Hi’s commute. For if hi [ Hi and hj [ Hj with i 2 j, then (hihj hi21)hj21 [ Hj hj 21 5 Hj and hi (hj hi21hj21) [ hiHi 5 Hi. Thus, hihj hi21hj 21 [ Hi > Hj 5 {e} (see Exercise 3), and, therefore, hi hj 5 hjhi. We next claim that each member of G can be expressed uniquely in the form h1h2 ? ? ? hn, where hi [ Hi. That there is at least one such representation is the content of condition 1 of the definition. To prove uniqueness, suppose that g 5 h1h2 ? ? ? hn and g 5 h19 h29 ? ? ? hn9, where hi and h9i belong to Hi for i 5 1, . . . , n. Then, using the fact that the h’s from different Hi’s commute, we can solve the equation h1h2 ? ? ? hn 5 h91 h92 ? ? ? h9n for h9n h n21 to obtain h9n hn21 5 (h91)21h1(h29)21h2 ? ? ? (h9n21)21hn21.

(1)

9 | Normal Subgroups and Factor Groups

191

But then h9nhn21 [ H1H2 ? ? ? Hn21 > Hn 5 {e}, so that h9n hn 21 5 e and, therefore, h9n 5 hn. At this point, we can cancel hn and h9n from opposite sides of the equal sign in Equation (1) and repeat the preceding argument to obtain hn21 5 hn21 9 . Continuing in this fashion, we eventually have hi 5 h9i for i 5 1, . . . , n. With our claim established, we may now define a function f from G to H1 % H2 % ? ? ? % Hn by f(h1h2 ? ? ? hn) 5 (h1, h2, . . . , hn). We leave to the reader the easy verification that f is an isomorphism. The next theorem provides an important application of Theorem 9.6. Theorem 9.7 Classification of Groups of Order p2 Every group of order p2, where p is a prime, is isomorphic to Zp2 or Zp % Zp.

PROOF Let G be a group of order p2 , where p is a prime. If G has an element of order p2 , then G is isomorphic to Z p2. So, by Corollary 2 of Lagrange’s Theorem, we may assume that every nonidentity element of G has order p. First we show that for any element a, the subgroup a is normal in G. If this is not the case then there is an element b in G such that bab21 is not in a. Then a and bab21  are distinct subgroups of order p. Since a x bab21  is a subgroup of both a and bab21  , we have that a x bab21  5 5e6 . From this it follows that the distinct left cosets of bab21  are bab21  , abab21  , a2 bab21  , . . . , ap21 bab21  . Since b21 must lie in one of these cosets, we may write b21 in the form b21 5 ai (bab21 ) j 5 aibajb21 for some i and j. Canceling the b21 terms, we obtain e 5 aibaj and therefore b 5 a2i2j [ a . This contradiction verifies our assertion that every subgroup of the form a is normal in G. To complete the proof, let x be any nonidentity element in G and y be any element of G not in x. Then, by comparing orders and using Theorem 9.6, we see that G 5 x 3 y < Z p % Z p . As an immediate corollary of Theorem 9.7, we have the following important fact. Corollary If G is a group of order p 2, where p is a prime, then G is Abelian.

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Groups

We mention in passing that if G 5 H1 % H2 % ? ? ? % Hn, then G can be expressed as the internal direct product of subgroups isomorphic to H1, H2, . . . , Hn. For example, if G 5 H1 % H2, then G 5 H1 3 H2, where H1 5 H1 % {e} and H2 5 {e} % H2. The topic of direct products is one in which notation and terminology vary widely. Many authors use H 3 K to denote both the internal direct product and the external direct product of H and K, making no notational distinction between the two products. A few authors define only the external direct product. Many people reserve the notation H % K for the situation where H and K are Abelian groups under addition and call it the direct sum of H and K. In fact, we will adopt this terminology in the section on rings (Part 3), since rings are always Abelian groups under addition. The U-groups provide a convenient way to illustrate the preceding ideas and to clarify the distinction between internal and external direct products. It follows directly from Theorem 8.3 and its corollary and Theorem 9.6 that if m 5 n1n2 ? ? ? nk, where gcd(ni, nj) 5 1 for i 2 j, then U(m) 5 Um/n1(m) 3 Um/n2(m) 3 ? ? ? 3 Um/nk(m) L U(n1) % U(n2) % ? ? ? % U(nk). Let us return to the examples given following Theorem 8.3. U(105) 5 U(15 ? 7) 5 U15(105) 3 U7(105) 5 {1, 16, 31, 46, 61, 76} 3 {1, 8, 22, 29, 43, 64, 71, 92} L U(7) % U(15), U(105) 5 U(5 ? 21) 5 U5(105) 3 U21(105) 5 {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101} 3 {1, 22, 43, 64} L U(21) % U(5), U(105) 5 U(3 ? 5 ? 7) 5 U35(105) 3 U21(105) 3 U15(105) 5 {1, 71} 3 {1, 22, 43, 64} 3 {1, 16, 31, 46, 61, 76} L U(3) % U(5) % U(7).

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Exercises Understanding is a kind of ecstasy. CARL SAGAN

1. Let H 5 {(1), (12)}. Is H normal in S3? 2. Prove that An is normal in Sn. 3. Show that if G is the internal direct product of H1, H2, . . . , Hn and i 2 j with 1 # i # n, 1 # j # n, then Hi > Hj 5 {e}. (This exercise is referred to in this chapter.) a b d ` a, b, d P R, ad 2 0 f . Is H a normal sub4. Let H 5 e c 0 d group of GL(2, R)? 5. Let G 5 GL(2, R) and let K be a subgroup of R*. Prove that H 5 {A [ G| det A [ K} is a normal subgroup of G. 6. Viewing 3 and 12 as subgroups of Z, prove that 3/12 is isomorphic to Z4. Similarly, prove that 8/48 is isomorphic to Z6. Generalize to arbitrary integers k and n. 7. Prove that if H has index 2 in G, then H is normal in G. (This exercise is referred to in Chapters 24 and 25 and this chapter.) 8. Let H 5 {(1), (12)(34)} in A4. a. Show that H is not normal in A4. b. Referring to the multiplication table for A4 in Table 5.1 on page 107, show that, although a6H 5 a7H and a9H 5 a11H, it is not true that a6a9H 5 a7a11H. Explain why this proves that the left cosets of H do not form a group under coset multiplication. 9. Let G 5 Z4 % U(4), H 5 (2, 3), and K 5 (2, 1). Show that G/H is not isomorphic to G/K. (This shows that H L K does not imply that G/H L G/K.) 10. Prove that a factor group of a cyclic group is cyclic. 11. Let H be a normal subgroup of G. If H and G/H are Abelian, must G be Abelian? 12. Prove that a factor group of an Abelian group is Abelian. 13. If H is a subgroup of G and a, b [ G, prove that (ab)H 5 a(bH). 14. What is the order of the element 14 1 8 in the factor group Z24/8? 15. What is the order of the element 4U5(105) in the factor group U(105)/U5(105)? 16. Recall that Z(D6) 5 {R0, R180}. What is the order of the element R60Z(D6) in the factor group D6/Z(D6)?

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Groups

17. 18. 19. 20. 21. 22. 23. 24. 25.

26.

27. 28.

29. 30. 31.

32. 33.

34. 35.

36.

Let G 5 Z/20 and H 5 4/20. List the elements of H and G/H. What is the order of the factor group Z60/15? What is the order of the factor group (Z10 % U(10))/(2, 9)? Construct the Cayley table for U(20)/U5(20). Prove that an Abelian group of order 33 is cyclic. Determine the order of (Z % Z)/(2, 2). Is the group cyclic? Determine the order of (Z % Z)/(4, 2). Is the group cyclic? The group (Z4 % Z12)/(2, 2) is isomorphic to one of Z8, Z4 % Z2, or Z2 % Z2 % Z2. Determine which one by elimination. Let G 5 U(32) and H 5 {1, 31}. The group G/H is isomorphic to one of Z8, Z4 % Z2, or Z2 % Z2 % Z2. Determine which one by elimination. Let G be the group of quarternions given by the table in Exercise 4 of the Supplementary Exercises for Chapters 1–4 on page 91, and let H be the subgroup {e, a2}. Is G/H isomorphic to Z4 or Z2 % Z2? Let G 5 U(16), H 5 {1, 15}, and K 5 {1, 9}. Are H and K isomorphic? Are G/H and G/K isomorphic? Let G 5 Z4 % Z4, H 5 {(0, 0), (2, 0), (0, 2), (2, 2)}, and K 5 (1, 2). Is G/H isomorphic to Z4 or Z2 % Z2? Is G/K isomorphic to Z4 or Z2 % Z2? Prove that A 4 % Z 3 has no subgroup of order 18. Express U(165) as an internal direct product of proper subgroups in four different ways. Let R* denote the group of all nonzero real numbers under multiplication. Let R1 denote the group of positive real numbers under multiplication. Prove that R* is the internal direct product of R1 and the subgroup {1, 21}. Prove that D4 cannot be expressed as an internal direct product of two proper subgroups. Let H and K be subgroups of a group G. If G 5 HK and g 5 hk, where h [ H and k [ K, is there any relationship among |g|, |h|, and |k|? What if G 5 H 3 K? In Z, let H 5 5 and K 5 7. Prove that Z 5 HK. Does Z 5 H 3 K? Let G 5 {3a6b10c | a, b, c [ Z} under multiplication and H 5 {3a6b12c | a, b, c [ Z} under multiplication. Prove that G 5 3 3 6 3 10, whereas H 2 3 3 6 3 12. Determine all subgroups of R* (nonzero reals under multiplication) of index 2.

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37. Let G be a finite group and let H be a normal subgroup of G. Prove that the order of the element gH in G/H must divide the order of g in G. 38. Let H be a normal subgroup of G and let a belong to G. If the element aH has order 3 in the group G/H and |H| 510, what are the possibilities for the order of a? 39. If H is a normal subgroup of a group G, prove that C(H), the centralizer of H in G, is a normal subgroup of G. 40. An element is called a square if it can be expressed in the form b2 for some b. Suppose that G is an Abelian group and H is a subgroup of G. If every element of H is a square and every element of G/H is a square, prove that every element of G is a square. Does your proof remain valid when “square” is replaced by “nth power,” where n is any integer? 41. Show, by example, that in a factor group G/H it can happen that aH 5 bH but |a| 2 |b|. (Do not use a 5 e or b 5 e.) 42. Observe from the table for A4 given in Table 5.1 on page 107 that the subgroup given in Example 6 of this chapter is the only subgroup of A4 of order 4. Why does this imply that this subgroup must be normal in A4? Generalize this to arbitrary finite groups. 43. Let p be a prime. Show that if H is a subgroup of a group of order 2p that is not normal, then H has order 2. 44. Show that D 13 is isomorphic to Inn(D 13 ) . 45. Suppose that N is a normal subgroup of a finite group G and H is a subgroup of G. If ZG>N Z is prime, prove that H is contained in N or that NH 5 G. 46. If G is a group and |G : Z(G)| 5 4 , prove that G>Z(G) < Z 2 % Z 2. 47. Suppose that G is a non-Abelian group of order p3, where p is a prime, and Z(G) 2 {e}. Prove that |Z(G)| 5 p. 48. If |G| 5 pq, where p and q are primes that are not necessarily distinct, prove that |Z(G)| 5 1 or pq. 49. Let N be a normal subgroup of G and let H be a subgroup of G. If N is a subgroup of H, prove that H/N is a normal subgroup of G/N if and only if H is a normal subgroup of G. 50. Let G be an Abelian group and let H be the subgroup consisting of all elements of G that have finite order (See Exercise 18 in the Supplementary Exercises for Chapters 1–4). Prove that every nonidentity element in G/H has infinite order. 51. Determine all subgroups of R* that have finite index.

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Groups

52. Let G 5 {61, 6i, 6j, 6k}, where i2 5 j2 5 k2 5 21, 2i 5 (21)i, 12 5 (21)2 5 1, ij 5 2ji 5 k, jk 5 2kj 5 i, and ki 5 2ik 5 j. a. Construct the Cayley table for G. b. Show that H 5 {1, 21} v G. c. Construct the Cayley table for G>H . Is G>H isomorphic to Z 4 or Z 2 % Z 2? (The rules involving i, j, and k can be remembered by using the circle below. i

k

53.

54. 55.

56. 57.

58.

j

Going clockwise, the product of two consecutive elements is the third one. The same is true for going counterclockwise, except that we obtain the negative of the third element.) This is the group of quaternions that was given in another form in Exercise 4 in the Supplementary Exercises for Chapters 1–4. It was invented by William Hamilton in 1843. The quaternions are used to describe rotations in three-dimensional space, and they are used in physics. The quaternions can be used to extend the complex numbers in a natural way. In D4, let K 5 {R0, D} and let L 5 {R0, D, D9, R180}. Show that K v L v D4, but that K is not normal in D4. (Normality is not transitive. Compare Exercise 4, Supplementary Exercises for Chapters 5–8.) Show that the intersection of two normal subgroups of G is a normal subgroup of G. Generalize. Let N be a normal subgroup of G and let H be any subgroup of G. Prove that NH is a subgroup of G. Give an example to show that NH need not be a subgroup of G if neither N nor H is normal. (This exercise is referred to in Chapter 24.) If N and M are normal subgroups of G, prove that NM is also a normal subgroup of G. Let N be a normal subgroup of a group G. If N is cyclic, prove that every subgroup of N is also normal in G. (This exercise is referred to in Chapter 24.) Without looking at inner automorphisms of Dn, determine the number of such automorphisms.

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59. Let H be a normal subgroup of a finite group G and let x [ G. If gcd(|x|, |G/H|) 5 1, show that x [ H. (This exercise is referred to in Chapter 25.) 60. Let G be a group and let G9 be the subgroup of G generated by the set S 5 {x21y21xy | x, y [ G}. (See Exercise 3, Supplementary Exercises for Chapters 5–8, for a more complete description of G9.) a. Prove that G9 is normal in G. b. Prove that G/G9 is Abelian. c. If G/N is Abelian, prove that G9 # N. d. Prove that if H is a subgroup of G and G9 # H, then H is normal in G. 61. If N is a normal subgroup of G and |G/N| 5 m, show that xm [ N for all x in G. 62. Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. 63. If G is non-Abelian, show that Aut(G) is not cyclic. 64. Let |G| 5 pnm, where p is prime and gcd( p, m) 5 1. Suppose that H is a normal subgroup of G of order pn. If K is a subgroup of G of order pk, show that K # H. 65. Suppose that H is a normal subgroup of a finite group G. If G/H has an element of order n, show that G has an element of order n. Show, by example, that the assumption that G is finite is necessary. (This exercise is referred to in this chapter.) 66. Recall that a subgroup N of a group G is called characteristic if f(N) 5 N for all automorphisms f of G. If N is a characteristic subgroup of G, show that N is a normal subgroup of G. 67. In D4, let _ 5 {R0, H}. Form an operation table for the cosets _, D_, V_, and D9_. Is the result a group table? Does your answer contradict Theorem 9.2? 68. Show that S4 has a unique subgroup of order 12. 69. If |G| 5 30 and |Z(G)| 5 5, what is the structure of G/Z(G)? 70. If H is a normal subgroup of G and |H| 5 2, prove that H is contained in the center of G. 71. Prove that A5 cannot have a normal subgroup of order 2. 72. Let G be a finite group and let H be an odd-order subgroup of G of index 2. Show that the product of all the elements of G (taken in any order) cannot belong to H.

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Groups

73. Let G be a group and p a prime. Suppose that H 5 5g p Zg [ G6 is a subgroup of G. Show that H is normal and that every nonidentity element of G>H has order p. 74. Suppose that H is a normal subgroup of G. If |H| 5 4 and gH has order 3 in G/H, find a subgroup of order 12 in G. 75. Let G be a group and H a subgroup of G of index 2. Show that H contains every element of G of odd order.

Suggested Readings Michael Brennan and Des MacHale, “Variations on a Theme: A4 Definitely Has No Subgroup of Order Six!,” Mathematics Magazine, 73 (2000): 36–40. The authors offer 11 proofs that A4 has no subgroup of order 6. These proofs provide a review of many of the ideas covered thus far in this text. J. A. Gallian, R. S. Johnson, and S. Peng. “On the Quotient Structure of Zn,” Pi Mu Epsilon Journal, 9 (1993): 524–526. The authors determine the structure of the group (Z % Z)/(a, b) and related groups. This article can be downloaded at http://www.d.umn .edu/~jgallian/quotient_structures.pdf Tony Rothman, “Genius and Biographers: The Fictionalization of Évariste Galois,” The American Mathematical Monthly 89 (1982): 84–106. The author convincingly argues that three of the most widely read accounts of Galois’ life are highly fictitious. Paul F. Zweifel, “Generalized Diatonic and Pentatonic Scales: A Grouptheoretic Approach,” Perspectives of New Music, 34 (1996): 140–161. The author discusses how group-theoretic notions such as subgroups, cosets, factor groups, and isomorphisms of Z12 and Z20 relate to musical scales, tuning, temperament, and structure.

Evariste Galois Galois at seventeen was making discoveries of epochal significance in the theory of equations, discoveries whose consequences are not yet exhausted after more than a century. E. T. BELL, Men of Mathematics

This French stamp was issued as part of the 1984 “Celebrity Series” in support of the Red Cross Fund.

ÉVARISTE GALOIS (pronounced gal-WAH) was born on October 25, 1811, near Paris. Although he had mastered the works of Legendre and Lagrange at age 15, Galois twice failed his entrance examination to l’Ecole Polytechnique. He did not know some basic mathematics, and he did mathematics almost entirely in his head, to the annoyance of the examiner. At 18, Galois wrote his important research on the theory of equations and submitted it to the French Academy of Sciences for publication. The paper was given to Cauchy for refereeing. Cauchy, impressed by the paper, agreed to present it to the academy, but he never did. At the age of 19, Galois entered a

paper of the highest quality in the competition for the Grand Prize in Mathematics, given by the French Academy of Sciences. The paper was given to Fourier, who died shortly thereafter. Galois’s paper was never seen again. Galois spent most of the last year and a half of his life in prison for revolutionary political offenses. While in prison, he attempted suicide and prophesied that he would die in a duel. On May 30, 1832, Galois was shot in a duel and died the next day at the age of 20. Among the many concepts introduced by Galois are normal subgroups, isomorphisms, simple groups, finite fields, and Galois theory. His work provided a method for disposing of several famous constructability problems, such as trisecting an arbitrary angle and doubling a cube. Galois’s entire collected works fill only 60 pages. To find more information about Galois, visit: http://www-groups.dcs .st-and.ac.uk/~history/

199

Group 10 Homomorphisms All modern theories of nuclear and electromagnetic interactions are based on group theory. ANDREW WATSON, New Scientist

Definition and Examples In this chapter, we consider one of the most fundamental ideas of algebra—homomorphisms. The term homomorphism comes from the Greek words homo, “like,” and morphe, “form.” We will see that a homomorphism is a natural generalization of an isomorphism and that there is an intimate connection between factor groups of a group and homomorphisms of a group. The concept of group homomorphisms was introduced by Camille Jordan in 1870, in his influential book Traité des Substitutions. Definition Group Homomorphism

A homomorphism f from a group G to a group G is a mapping from G into G that preserves the group operation; that is, f(ab) 5 f(a)f(b) for all a, b in G.

Before giving examples and stating numerous properties of homomorphisms, it is convenient to introduce an important subgroup that is intimately related to the image of a homomorphism. (See property 4 of Theorem 10.1.) Definition Kernel of a Homomorphism

The kernel of a homomorphism f from a group G to a group with identity e is the set {x [ G | f(x) 5 e}. The kernel of f is denoted by Ker f.

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EXAMPLE 1 Any isomorphism is a homomorphism that is also onto and one-to-one. The kernel of an isomorphism is the trivial subgroup. EXAMPLE 2 Let R* be the group of nonzero real numbers under multiplication. Then the determinant mapping A → det A is a homomorphism from GL(2, R) to R*. The kernel of the determinant mapping is SL(2, R). EXAMPLE 3 The mapping f from R* to R*, defined by f(x) 5 |x|, is a homomorphism with Ker f 5 {1, 21}. EXAMPLE 4 Let R[x] denote the group of all polynomials with real coefficients under addition. For any f in R[x], let f9 denote the derivative of f. Then the mapping f S f9 is a homomorphism from R[x] to itself. The kernel of the derivative mapping is the set of all constant polynomials. EXAMPLE 5 The mapping f from Z to Zn, defined by f(m) 5 m mod n, is a homomorphism (see Exercise 11 in Chapter 0). The kernel of this mapping is n. EXAMPLE 6 The mapping f(x) 5 x2 from R*, the nonzero real numbers under multiplication, to itself is a homomorphism, since f(ab) 5 (ab)2 5 a2b2 5 f(a)f(b) for all a and b in R*. (See Exercise 5.) The kernel is {1, –1}. EXAMPLE 7 The mapping f(x) 5 x2 from R, the real numbers under addition, to itself is not a homomorphism, since f(a 1 b) 5 (a 1 b)2 5 a2 1 2ab 1 b2, whereas f(a) 1 f(b) 5 a2 1 b2. When defining a homomorphism from a group in which there are several ways to represent the elements, caution must be exercised to ensure that the correspondence is a function. (The term well-defined is often used in this context.) For example, since 3(x 1 y) 5 3x 1 3y in Z6, one might believe that the correspondence x 1 3 S 3x from Z/3 to Z6 is a homomorphism. But it is not a function, since 0 1 3 5 3 1 3 in Z/3 but 3 ? 0 2 3 ? 3 in Z6. For students who have had linear algebra, we remark that every linear transformation is a group homomorphism and the nullspace is the same as the kernel. An invertible linear transformation is a group isomorphism.

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Properties of Homomorphisms Theorem 10.1 Properties of Elements Under Homomorphisms Let f be a homomorphism from a group G to a group G and let g be an element of G. Then 1. f carries the identity of G to the identity of G. 2. f(gn) 5 (f(g))n for all n in Z. 3. If |g| is finite, then |f(g)| divides |g|. 4. Ker f is a subgroup of G. 5. f(a) 5 f(b) if and only if aKer f 5 bKer f. 6. If f(g) 5 g9, then f21(g9) 5 {x [ G | f(x) 5 g9} 5 gKer f.

PROOF The proofs of properties 1 and 2 are identical to the proofs of properties 1 and 2 of isomorphisms in Theorem 6.2. To prove property 3, notice that properties 1 and 2 together with gn 5 e imply that e 5 f(e) 5 f(gn) 5 (f(g))n. So, by Corollary 2 to Theorem 4.1, we have |f(g)| divides n. By property 1 we know that Ker f is not empty. So, to prove property 4, we assume that a, b [ Ker f and show that ab21 [ Ker f. Since f(a) 5 e and f(b) 5 e, we have f(ab21) 5 f(a)f(b21) 5 f(a)(f(b))21 5 ee21 5 e. So, ab21 [ Ker f. To prove property 5, first assume that f(a) 5 f(b). Then e 5 (f(b))21f(a) 5 f(b21)f(a) 5 f(b21a), so that b21a[ Ker f. It now follows from property 5 of the lemma in Chapter 7 that bKer f 5 aKer f. Reversing this argument completes the proof. To prove property 6, we must show that f21(g9) # gKer f and that gKer f # f21(g9). For the first inclusion, let x [ f21(g9), so that f(x) 5 g9. Then f(g) 5 f(x) and by property 5 we have gKer f 5 xKer f and therefore x [ gKer f. This completes the proof that f21(g9) # gKer f. To prove that gKer f # f21(g9), suppose that k [ Ker f. Then f(gk) 5 f(g)f(k) 5 g9e 5 g9. Thus, by definition, gk [ f21(g9). Since homomorphisms preserve the group operation, it should not be a surprise that they preserve many group properties.

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Theorem 10.2 Properties of Subgroups Under Homomorphisms Let f be a homomorphism from a group G to a group G and let H be a subgroup of G. Then 1. f(H) 5 {f(h) | h [ H} is a subgroup of G . 2. If H is cyclic, then f(H) is cyclic. 3. f H is Abelian, then f(H) is Abelian. 4. If H is normal in G, then f(H) is normal in f(G). 5. If |Ker f| 5 n, then f is an n-to-1 mapping from G onto f(G). 6. If |H| 5 n, then |f(H)| divides n. 7. If K is a subgroup of G , then f21(K ) 5 {k [ G | f(k) [ K } is a subgroup of G. 8. If K is a normal subgroup of G , then f21(K ) 5 {k [ G | f(k) [ K } is a normal subgroup of G. 9. If f is onto and Ker f 5 {e}, then f is an isomorphism from G to G .

PROOF First note that the proofs of properties 1, 2, and 3 are identical to the proofs of properties 4, 3, and 2, respectively, of Theorem 6.3, since those proofs use only the fact that an isomorphism is an operation-preserving mapping. To prove property 4, let f(h) [ f(H) and f(g) [ f(G). Then f(g)f(h)f(g)21 5 f(ghg21) [ f(H), since H is normal in G. Property 5 follows directly from property 6 of Theorem 10.1 and the fact that all cosets of Ker f 5 f21(e) have the same number of elements. To prove property 6, let fH denote the restriction of f to the elements of H. Then fH is a homomorphism from H onto f(H). Suppose |Ker fH| 5 t. Then, by property 5, fH is a t-to-1 mapping. So, |f(H)|t 5 |H|. To prove property 7, we use the One-Step Subgroup Test. Clearly, e [ f21(K ), so that f21(K ) is not empty. Let k1, k2 [ f21(K ). Then, by the definition of f21(K ), we know that f(k1), f(k2) [ K . Thus, f(k2)21 [ K as well and f(k1k221) 5 f(k1)f(k2)21 [ K . So, by definition of f21(K ), we have k1k221 [ f21(K ). To prove property 8, we use the normality test given in Theorem 9.1. Note that every element in xf21(K )x21 has the form xkx21, where f(k) [ K . Thus, since K is normal in G, f(xkx21) 5 f(x)f(k)(f(x))21 [ K , and, therefore, xkx21 [ f21(K ). Finally, property 9 follows directly from property 5.

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A few remarks about Theorems 10.1 and 10.2 are in order. Students should remember the various properties of these theorems in words. For example, properties 2 and 3 of Theorem 10.2 say that the homomorphic image of a cyclic group is cyclic and the homomorphic image of an Abelian group is Abelian. Property 4 of Theorem 10.2 says that the homomorphic image of a normal subgroup of G is normal in the image of G. Property 5 of Theorem 10.2 says that if f is a homomorphism from G to G, then every element of G that gets “hit” by f gets hit the same number of times as does the identity. The set f21(g9) defined in property 6 of Theorem 10.1 is called the inverse image of g9 (or the pullback of g9). Note that the inverse image of an element is a coset of the kernel and that every element in that coset has the same image. Similarly, the set f21(K ) defined in property 7 of Theorem 10.2 is called the inverse image of K (or the pullback of K ). Property 6 of Theorem 10.1 is reminiscent of something from linear algebra and differential equations. Recall that if x is a particular solution to a system of linear equations and S is the entire solution set of the corresponding homogeneous system of linear equations, then x 1 S is the entire solution set of the nonhomogeneous system. In reality, this statement is just a special case of property 6. Properties 1 and 6 of Theorem 10.1 and property 5 of Theorem 10.2 are pictorially represented in Figure 10.1. The special case of property 8 of Theorem 10.2, where K 5 {e}, is of such importance that we single it out. Corollary Kernels Are Normal Let f be a group homomorphism from G to G . Then Ker f is a normal subgroup of G.

The next two examples illustrate several properties of Theorems 10.1 and 10.2. EXAMPLE 8 Consider the mapping f from C* to C* given by f(x) 5 x4. Since (xy)4 5 x4y4, f is a homomorphism. Clearly, Ker f 5 {x | x4 5 1} 5 {1, 21, i, 2i}. So, by property 5 of Theorem 10.2, we know that f is a 4-to-1 mapping. Now let’s find all elements 4 that map to, say, 2. Certainly, f(" 2 ) 5 2. Then, by property 6 of 4 Theorem 10.1, the set of all elements that map to 2 is " 2 Ker f 5 {"4 2 , 2"4 2 , "4 2 i, 2"4 2 i}.

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G

G

Ker φ = φ 21(e)

gKer φ = φ 21(g9)

e = g1, g2,..., gn

g, gg2,..., ggn

φ

φ

φ (G) e

φ (g) = g9

Figure 10.1

Finally, we verify a specific instance of property 3 of Theorem 10.1 and of property 2 and property 6 of Theorem 10.2. Let H 5 cos 30° 1 i sin 30°. It follows from DeMoivre’s Theorem (Example 7 in Chapter 0) that |H| 5 12, f(H) 5 cos 120° 1 i sin 120°, and |f(H)| 5 3. EXAMPLE 9 Define f:Z12 → Z12 by f(x) 5 3x. To verify that f is a homomorphism, we observe that in Z12, 3(a 1 b) 5 3a 1 3b (since the group operation is addition modulo 12). Direct calculations show that Ker f 5 {0, 4, 8}. Thus, we know from property 5 of Theorem 10.2 that f is a 3-to-1 mapping. Since f(2) 5 6, we have by property 6 of Theorem 10.1 that f21(6) 5 2 1 Ker f 5 {2, 6, 10}. Notice also that 2 is cyclic and f(2) 5 {0, 6} is cyclic. Moreover, |2| 5 6 and |f(2)| 5 |6| 5 2, so |f(2)| divides |2| in agreement with property 3 of Theorem 10.1. Letting K5 {0, 6}, we see that the subgroup f21(K) 5 {0, 2, 4, 6, 8, 10}. This verifies property 7 of Theorem 10.2 in this particular case. The next example illustrates how one can easily determine all homomorphisms from a cyclic group to a cyclic group. EXAMPLE 10 We determine all homomorphisms from Z12 to Z30. By property 2 of Theorem 10.1, such a homomorphism is completely specified by the image of 1. That is, if 1 maps to a, then x maps to xa. Lagrange’s Theorem and property 3 of Theorem 10.1 require that |a| divide both 12 and 30. So, |a| 5 1, 2, 3, or 6. Thus, a 5 0, 15, 10, 20, 5, or 25. This gives us a list of candidates for the homomorphisms. That each of these six possibilities yields an operation-preserving, welldefined function can now be verified by direct calculations. [Note that gcd(12, 30) 5 6. This is not a coincidence!]

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) 3 2

) (1

3

)

(2

2 (1

) 2 3 (1

) 3 2 (1 (1

) (1 ) (1 2 3 )

(1

(1

)

(1

2

(1

)

3

(1

)

2

)

(1

3

)

(2

3

)

(2

3

)

EXAMPLE 11 The mapping from Sn to Z2 that takes an even permutation to 0 and an odd permutation to 1 is a homomorphism. Figure 10.2 illustrates the telescoping nature of the mapping.

O O

O

1

1

O

Figure 10.2 Homomorphism from S3 to Z2.

The First Isomorphism Theorem In Chapter 9, we showed that for a group G and a normal subgroup H, we could arrange the Cayley table of G into boxes that represented the cosets of H in G, and that these boxes then became a Cayley table for G/H. The next theorem shows that for any homomorphism f of G and the normal subgroup Ker f, the same process produces a Cayley table isomorphic to the homomorphic image of G. Thus, homomorphisms, like factor groups, cause a systematic collapse of a group to a simpler but closely related group. This can be likened to viewing a group through the reverse end of a telescope—the general features of the group are present, but the apparent size is diminished. The important

207

10 | Group Homomorphisms

relationship between homomorphisms and factor groups given below is often called the Fundamental Theorem of Group Homomorphisms. Theorem 10.3 First Isomorphism Theorem (Jordan, 1870) Let f be a group homomorphism from G to G . Then the mapping from G/Ker f to f(G), given by gKer f → f(g), is an isomorphism. In symbols, G/Ker f L f(G).

PROOF Let us use c to denote the correspondence gKer f S f(g). That c is well defined (that is, the correspondence is independent of the particular coset representative chosen) and one-to-one follows directly from property 5 of Theorem 10.1. To show that c is operationpreserving, observe that c(xKer f yKer f) 5 c(xyKer f) 5 f(xy) 5 f(x) f(y) 5 c(xKer f)c(yKer f). The next corollary follows directly from Theorem 10.3, property 1 of Theorem 10.2, and Lagrange’s Theorem. Corollary If f is a homomorphism from a finite group G to G , then |f(G)| divides |G| and |G |.

EXAMPLE 10 To illustrate Theorem 10.3 and its proof, consider the homomorphism f from D4 to itself given by R0

R180 R0

R90

R270

H

H

V R180

D

D9 V

Then Ker f 5 {R0, R180}, and the mapping c in Theorem 10.3 is R0Ker f S R0, R90Ker f S H, HKer f S R180, DKer f S V. It is straight-forward to verify that the mapping c is an isomorphism. Mathematicians often give a pictorial representation of Theorem 10.3, as follows: φ

G γ

φ (G) ψ

G/Ker φ

208

Groups

where g:G S G/Ker f is defined as g(g) 5 gKer f. The mapping g is called the natural mapping from G to G/Ker f. Our proof of Theorem 10.3 shows that cg 5 f. In this case, one says that the preceding diagram is commutative. As a consequence of Theorem 10.3, we see that all homomorphic images of G can be determined using G. We may simply consider the various factor groups of G. For example, we know that the homomorphic image of an Abelian group is Abelian because the factor group of an Abelian group is Abelian. We know that the number of homomorphic images of a cyclic group G of order n is the number of divisors of n, since there is exactly one subgroup of G (and therefore one factor group of G) for each divisor of n. (Be careful: The number of homomorphisms of a cyclic group of order n need not be the same as the number of divisors of n, since different homomorphisms can have the same image.) An appreciation for Theorem 10.3 can be gained by looking at a few examples. EXAMPLE 13 Z/8N9 L ZN Consider the mapping from Z to Zn defined in Example 5. Clearly, its kernel is n. So, by Theorem 10.3, Z/n L Zn. EXAMPLE 14 The Wrapping Function Recall the wrapping function W from trigonometry. The real number line is wrapped around a unit circle in the plane centered at (0, 0) with the number 0 on the number line at the point (1, 0), the positive reals in the counterclockwise direction and the negative reals in the clockwise direction (see Figure 10.3). The function W assigns to each real number a the point a radians from (1, 0) on the circle. This mapping is a homomorphism from the group R under addition onto the circle group (the group of complex numbers of magnitude 1 under multiplication). Indeed, it follows from elementary facts of trigonometry that W(x) 5 cos x 1 i sin x and W(x 1 y) 5 W(x)W(y). Since W is periodic of period 2p, Ker W 5 2p. So, from the First Isomorphism Theorem, we see that R/2p is isomorphic to the circle group. W(2)

W(3)

W(1) (1, 0) W(0)

(0, 0) W(21)

Figure 10.3

10 | Group Homomorphisms

209

Our next example is a theorem that is used repeatedly in Chapters 24 and 25. EXAMPLE 15 The N/C Theorem Let H be a subgroup of a group G. Recall that the normalizer of H in G is N(H) 5 {x [ G | xHx21 5 H} and the centralizer of H in G is C(H) 5 {x [ G | xhx21 5 h for all h in H}. Consider the mapping from N(H) to Aut(H) given by g S fg, where fg is the inner automorphism of H induced by g [that is, fg(h) 5 ghg21 for all h in H]. This mapping is a homomorphism with kernel C(H). So, by Theorem 10.3, N(H)/C(H) is isomorphic to a subgroup of Aut(H). As an application of the N/C Theorem, we will show that every group of order 35 is cyclic. EXAMPLE 16 Let G be a group of order 35. By Lagrange’s Theorem, every nonidentity element of G has order 5, 7, or 35. If some element has order 35, G is cyclic. So we may assume that all nonidentity elements have order 5 or 7. However, not all such elements can have order 5, since elements of order 5 come 4 at a time (if |x| 5 5, then |x2| 5 |x3| 5 |x4| 5 5) and 4 does not divide 34. Similarly, since 6 does not divide 34, not all nonidentity elements can have order 7. So, G has elements of order 7 and order 5. Since G has an element of order 7, it has a subgroup of order 7. Let us call it H. In fact, H is the only subgroup of G of order 7, for if K is another subgroup of G of order 7, we have by Exercise 7 of the Supplementary Exercises for Chapters 5–8 that |HK| 5 |H||K|/|H > K| 5 7 ? 7/1 5 49. But, of course, this is impossible in a group of order 35. Since for every a in G, aHa21 is also a subgroup of G of order 7 (see Exercise 1 of the Supplementary Exercises for Chapters 1–4), we must have aHa21 5 H. So, N(H) 5 G. Since H has prime order, it is cyclic and therefore Abelian. In particular, C(H) contains H. So, 7 divides |C(H)| and |C(H)| divides 35. It follows, then, that C(H) 5 G or C(H) 5 H. If C(H) 5 G, then we may obtain an element x of order 35 by letting x 5 hk, where h is a nonidentity element of H and k has order 5. On the other hand, if C(H) 5 H, then |C(H)| 5 7 and |N(H)/C(H)| 5 35/7 5 5. However, 5 does not divide |Aut(H)| 5 |Aut(Z7)| 5 6. This contradiction shows that G is cyclic. The corollary of Theorem 10.2 says that the kernel of every homomorphism of a group is a normal subgroup of the group. We conclude this chapter by verifying that the converse of this statement is also true.

210

Groups

Theorem 10.4 Normal Subgroups Are Kernels Every normal subgroup of a group G is the kernel of a homomorphism of G. In particular, a normal subgroup N is the kernel of the mapping g S gN from G to G/N.

PROOF Define g:G S G/N by g(g) 5 gN. (This mapping is called the natural homomorphism from G to G/N.) Then, g(xy) 5 (xy)N 5 xNyN 5 g(x)g(y). Moreover, g [ Ker g if and only if gN 5 g(g) 5 N, which is true if and only if g [ N (see property 2 of the lemma in Chapter 7). Examples 13, 14, and 15 illustrate the utility of the First Isomorphism Theorem. But what about homomorphisms in general? Why would one care to study a homomorphism of a group? The answer is that, just as was the case with factor groups of a group, homomorphic images of a group tell us some of the properties of the original group. One measure of the likeness of a group and its homomorphic image is the size of the kernel. If the kernel of the homomorphism of group G is the identity, then the image of G tells us everything (group theoretically) about G (the two being isomorphic). On the other hand, if the kernel of the homomorphism is G itself, then the image tells us nothing about G. Between these two extremes, some information about G is preserved and some is lost. The utility of a particular homomorphism lies in its ability to preserve the group properties we want, while losing some inessential ones. In this way, we have replaced G by a group less complicated (and therefore easier to study) than G; but, in the process, we have saved enough information to answer questions that we have about G itself. For example, if G is a group of order 60 and G has a homomorphic image of order 12 that is cyclic, then we know from properties 5, 7, and 8 of Theorem 10.2 that G has normal subgroups of orders 5, 10, 15, 20, 30, and 60. To illustrate further, suppose we are asked to find an infinite group that is the union of three proper subgroups. Instead of attempting to do this directly, we first make the problem easier by finding a finite group that is the union of three proper subgroups. Observing that Z2 % Z2 is the union of H1 5 1, 0, H2 5 0, 1, and H3 5 1, 1, we have found our finite group. Now all we need do is think of an infinite group that has Z2 % Z2 as a homomorphic image and pull back H1, H2, and H3, and our original problem is solved. Clearly, the mapping from Z2 % Z2 % Z onto Z2 % Z2 given by f(a, b, c) 5 (a, b) is such a mapping, and therefore Z2 % Z2 % Z is the union of f21(H1) 5 {(a, 0, c,) | a [ Z2, c [ Z}, f21(H2) 5 {(0, b, c) | b [ Z2, c [ Z}, and f21(H3) 5 {(a, a, c) | a [ Z2, c [ Z}.

10 | Group Homomorphisms

211

Although an isomorphism is a special case of a homomorphism, the two concepts have entirely different roles. Whereas isomorphisms allow us to look at a group in an alternative way, homomorphisms act as investigative tools. The following analogy between homomorphisms and photography may be instructive.† A photograph of a person cannot tell us the person’s exact height, weight, or age. Nevertheless, we may be able to decide from a photograph whether the person is tall or short, heavy or thin, old or young, male or female. In the same way, a homomorphic image of a group gives us some information about the group. In certain branches of group theory, and especially in physics and chemistry, one often wants to know all homomorphic images of a group that are matrix groups over the complex numbers (these are called group representations). Here, we may carry our analogy with photography one step further by saying that this is like wanting photographs of a person from many different angles (front view, profile, head-to-toe view, closeup, etc.), as well as x-rays! Just as this composite information from the photographs reveals much about the person, several homomorphic images of a group reveal much about the group.

Exercises The greater the difficulty, the more glory in surmounting it. Skillful pilots gain their reputation from storms and tempests. EPICURUS

1. 2. 3. 4. 5.

Prove that the mapping given in Example 2 is a homomorphism. Prove that the mapping given in Example 3 is a homomorphism. Prove that the mapping given in Example 4 is a homomorphism. Prove that the mapping given in Example 11 is a homomorphism. Let R* be the group of nonzero real numbers under multiplication, and let r be a positive integer. Show that the mapping that takes x to xr is a homomorphism from R* to R* and determine the kernel. Which values of r yield an isomorphism? 6. Let G be the group of all polynomials with real coefficients under addition. For each f in G, let f denote the antiderivative of f that passes through the point (0, 0). Show that the mapping f S f from G to G is a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if f denotes the antiderivative of f that passes through (0, 1)?

†All

perception of truth is the detection of an analogy. Henry David Thoreau, Journal.

212

Groups

7. If f is a homomorphism from G to H and s is a homomorphism from H to K, show that sf is a homomorphism from G to K. How are Ker f and Ker sf related? If f and s are onto and G is finite, describe [Ker sf : Ker f] in terms of |H| and |K|. 8. Let G be a group of permutations. For each s in G, define sgn(s) 5 e

11 if s is an even permutation, 21 if s is an odd permutation.

Prove that sgn is a homomorphism from G to the multiplicative group {11, 21}. What is the kernel? Why does this homomorphism allow you to conclude that An is a normal subgroup of S n of index 2? 9. Prove that the mapping from G % H to G given by (g, h) S g is a homomorphism. What is the kernel? This mapping is called the projection of G % H onto G. 10. Let G be a subgroup of some dihedral group. For each x in G, define f(x) 5 e

11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

11 21

if x is a rotation, if x is a reflection.

Prove that f is a homomorphism from G to the multiplicative group 511,216 . What is the kernel? Prove that (Z % Z)/((a, 0) 3 (0, b)) is isomorphic to Za % Zb. Suppose that k is a divisor of n. Prove that Zn/k L Zk. Prove that (A % B)/(A % {e}) L B. Explain why the correspondence x → 3x from Z12 to Z10 is not a homomorphism. Suppose that f is a homomorphism from Z30 to Z30 and Ker f 5 {0, 10, 20}. If f(23) 5 9, determine all elements that map to 9. Prove that there is no homomorphism from Z8 % Z2 onto Z4 % Z4. Prove that there is no homomorphism from Z16 % Z2 onto Z4 % Z4. Can there be a homomorphism from Z4 % Z4 onto Z8? Can there be a homomorphism from Z16 onto Z2 % Z2? Explain your answers. Suppose that there is a homomorphism f from Z17 to some group and that f is not one-to-one. Determine f. How many homomorphisms are there from Z20 onto Z8? How many are there to Z8? If f is a homomorphism from Z30 onto a group of order 5, determine the kernel of f.

10 | Group Homomorphisms

213

22. Suppose that f is a homomorphism from a finite group G onto G and that G has an element of order 8. Prove that G has an element of order 8. Generalize. 23. Suppose that f is a homomorphism from Z36 to a group of order 24. a. Determine the possible homomorphic images. b. For each image in part a, determine the corresponding kernel of f. 24. Suppose that f: Z50 S Z15 is a group homomorphism with f(7) 5 6. a. Determine f(x). b. Determine the image of f. c. Determine the kernel of f. d. Determine f21(3). That is, determine the set of all elements that map to 3. 25. How many homomorphisms are there from Z20 onto Z10? How many are there to Z10? 26. Determine all homomorphisms from Z4 to Z2 % Z2. 27. Determine all homomorphisms from Zn to itself. 28. Suppose that f is a homomorphism from S4 onto Z2. Determine Ker f. Determine all homomorphisms from S4 to Z2. 29. Suppose that there is a homomorphism from a finite group G onto Z10. Prove that G has normal subgroups of indexes 2 and 5. 30. Suppose that f is a homomorphism from a group G onto Z6 % Z2 and that the kernel of f has order 5. Explain why G must have normal subgroups of orders 5, 10, 15, 20, 30, and 60. 31. Suppose that f is a homomorphism from U(30) to U(30) and that Ker f 5 {1, 11}. If f(7) 5 7, find all elements of U(30) that map to 7. 32. Find a homomorphism f from U(30) to U(30) with kernel {1, 11} and f(7) 5 7. 33. Suppose that f is a homomorphism from U(40) to U(40) and that Ker f 5 {1, 9, 17, 33}. If f(11) 5 11, find all elements of U(40) that map to 11. 34. Find a homomorphism f from U(40) to U(40) with kernel {1, 9, 17, 33} and f(11) 5 11. 35. Prove that the mapping f: Z % Z S Z given by (a, b) S a 2 b is a homomorphism. What is the kernel of f? Describe the set f21(3) (that is, all elements that map to 3). 36. Suppose that there is a homomorphism f from Z % Z to a group G such that f((3, 2)) 5 a and f((2, 1)) 5 b. Determine f((4, 4)) in terms of a and b. Assume that the operation of G is addition.

214

Groups

37. Prove that the mapping x S x6 from C* to C* is a homomorphism. What is the kernel? 38. For each pair of positive integers m and n, we can define a homomorphism from Z to Zm % Zn by x S (x mod m, x mod n). What is the kernel when (m, n) 5 (3, 4)? What is the kernel when (m, n) 5 (6, 4)? Generalize. 39. (Second Isomorphism Theorem) If K is a subgroup of G and N is a normal subgroup of G, prove that K/(K > N) is isomorphic to KN/N. 40. (Third Isomorphism Theorem) If M and N are normal subgroups of G and N # M, prove that (G/N)/(M/N) L G/M. 41. Let f(d) denote the Euler phi function of d (see page 79). Show that the number of homomorphisms from Zn to Zk is Sf(d), where the sum runs over all common divisors d of n and k. [It follows from number theory that this sum is actually gcd(n, k).] 42. Let k be a divisor of n. Consider the homomorphism from U(n) to U(k) given by x S x mod k. What is the relationship between this homomorphism and the subgroup Uk(n) of U(n)? 43. Determine all homomorphic images of D4 (up to isomorphism). 44. Let N be a normal subgroup of a finite group G. Use the theorems of this chapter to prove that the order of the group element gN in G/N divides the order of g. 45. Suppose that G is a finite group and that Z10 is a homomorphic image of G. What can we say about |G|? Generalize. 46. Suppose that Z10 and Z15 are both homomorphic images of a finite group G. What can be said about |G|? Generalize. 47. Suppose that for each prime p, Zp is the homomorphic image of a group G. What can we say about |G|? Give an example of such a group. 48. (For students who have had linear algebra.) Suppose that x is a particular solution to a system of linear equations and that S is the entire solution set of the corresponding homogeneous system of linear equations. Explain why property 6 of Theorem 10.1 guarantees that x 1 S is the entire solution set of the nonhomogeneous system. In particular, describe the relevant groups and the homomorphism between them. 49. Let N be a normal subgroup of a group G. Use property 7 of Theorem 10.2 to prove that every subgroup of G/N has the form H/N, where H is a subgroup of G. (This exercise is referred to in Chapter 24.)

10 | Group Homomorphisms

215

50. Show that a homomorphism defined on a cyclic group is completely determined by its action on a generator of the group. 51. Use the First Isomorphism Theorem to prove Theorem 9.4. 52. Let a and b be group homomorphisms from G to G and let H 5 {g [ G | a(g) 5 b(g)}. Prove or disprove that H is a subgroup of G. 53. Let Z[x] be the group of polynomials in x with integer coefficients under addition. Prove that the mapping from Z[x] into Z given by f(x) S f(3) is a homomorphism. Give a geometric description of the kernel of this homomorphism. Generalize. 54. Prove that the mapping from R under addition to GL(2, R) that takes x to cos x sin x c d 2sin x cos x 55. 56. 57.

58.

59. 60. 61.

62.

is a group homomorphism. What is the kernel of the homomorphism? Suppose there is a homomorphism f from G onto Z2 % Z2. Prove that G is the union of three proper normal subgroups. If H and K are normal subgroups of G and H > K 5 {e}, prove that G is isomorphic to a subgroup of G/H % G/K. Suppose that H and K are distinct subgroups of G of index 2. Prove that H > K is a normal subgroup of G of index 4 and that G/(H > K) is not cyclic. Suppose that the number of homomorphisms from G to H is n. How many homomorphisms are there from G to H % H % ? ? ? % H (s terms)? When H is Abelian, how many homomorphisms are there from G % G % ? ? ? % G (s terms) to H? Prove that every group of order 77 is cyclic. Determine all homomorphisms from Z onto S3. Determine all homomorphisms from Z to S3. Suppose G is an Abelian group under addition with the property that for every positive integer n the set nG 5{ng|g [ G} 5 G. Show that every proper subgroup of G is properly contained in a proper subgroup of G. Name two familiar groups that satisfy the hypothesis. Let p be a prime. Determine the number of homomorphisms from Zp % Zp into Zp.

216

Groups

Computer Exercise A computer lets you make more mistakes faster than any invention in human history—with the possible exceptions of handguns and tequila. MITCH RATLIFFE

Software for the computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines the homomorphisms from Zm to Zn. (Recall that a homomorphism from Zm is completely determined by the image of 1.) Run the program for m 5 20 with various choices for n. Run the program for m 5 15 with various choices for n. What relationship do you see between m and n and the number of homomorphisms from Zm to Zn? For each choice of m and n, observe the smallest positive image of 1. Try to see the relationship between this image and the values of m and n. What relationship do you see between the smallest positive image of 1 and the other images of 1? Test your conclusions with other choices of m and n.

Camille Jordan Although these contributions [to analysis and topology] would have been enough to rank Jordan very high among his mathematical contemporaries, it is chiefly as an algebraist that he reached celebrity when he was barely thirty; and during the next forty years he was universally regarded as the undisputed master of group theory. J. DIEUDONNÉ, Dictionary of

Scientific Biography

CAMILLE JORDAN was born into a well-to-do family on January 5, 1838, in Lyons, France. Like his father, he graduated from the École Polytechnique and became an engineer. Nearly all of his 120 research papers in mathematics were written before his retirement from engineering in 1885. From 1873 until 1912, Jordan taught simultaneously at the École Polytechnique and at the College of France. In the great French tradition, Jordan was a universal mathematician who published in nearly every branch of mathematics. Among the concepts named after him are the Jordan canonical form in matrix theory, the Jordan curve theorem from topology, and the Jordan-Hölder theorem from group theory.

His classic book Traité des Substitutions, published in 1870, was the first to be devoted solely to group theory and its applications to other branches of mathematics. Another book that had great influence and set a new standard for rigor was his Cours d’analyse. This book gave the first clear definitions of the notions of volume and multiple integral. Nearly 100 years after this book appeared, the distinguished mathematician and mathematical historian B. L. van der Waerden wrote, “For me, every single chapter of the Cours d’analyse is a pleasure to read.” Jordan died in Paris on January 22, 1922. To find more information about Jordan, visit: http://www-groups.dcs .st-and.ac.uk/~history/

217

Fundamental 11 Theorem of Finite Abelian Groups By a small sample we may judge of the whole piece. MIGUEL DE CERVANTES, Don Quixote

The Fundamental Theorem In this chapter, we present a theorem that describes to an algebraist’s eye (that is, up to isomorphism) all finite Abelian groups in a standardized way. Before giving the proof, which is long and difficult, we discuss some consequences of the theorem and its proof. The first proof of the theorem was given by Leopold Kronecker in 1858. Theorem 11.1 Fundamental Theorem of Finite Abelian Groups Every finite Abelian group is a direct product of cyclic groups of prime-power order. Moreover, the number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.

Since a cyclic group of order n is isomorphic to Zn, Theorem 11.1 shows that every finite Abelian group G is isomorphic to a group of the form Zp1n1 % Zp2n2 % ? ? ? % Z pknk, where the pi’s are not necessarily distinct primes and the primepowers p1n1, p2n2, . . . , pknk are uniquely determined by G. Writing a group in this form is called determining the isomorphism class of G.

The Isomorphism Classes of Abelian Groups The Fundamental Theorem is extremely powerful. As an application, we can use it as an algorithm for constructing all Abelian groups of any order. Let’s look at groups whose orders have the form pk, where p is 218

11 | Fundamental Theorem of Finite Abelian Groups

219

prime and k # 4. In general, there is one group of order pk for each set of positive integers whose sum is k (such a set is called a partition of k); that is, if k can be written as k 5 n1 1 n2 1 ? ? ? 1 nt, where each ni is a positive integer, then Zp n1 % Zp n2 % ? ? ? % Zp nt is an Abelian group of order pk. Partitions of k 1 2

Possible direct products for G Zp Zp2

p3

111 3

Zp % Zp Zp3

p4

211 11111 4

Zp2 % Zp Zp % Zp % Zp Zp4

311 212

Zp3 % Zp Zp2 % Zp2

21111 1111111

Zp2 % Zp % Zp Zp % Zp % Zp % Zp

Order of G p p2

Furthermore, the uniqueness portion of the Fundamental Theorem guarantees that distinct partitions of k yield distinct isomorphism classes. Thus, for example, Z9 % Z3 is not isomorphic to Z3 % Z3 % Z3. A reliable mnemonic for comparing external direct products is the cancellation property: If A is finite, then A%BLA%C

if and only if

BLC

(see [1]).

Thus Z4 % Z4 is not isomorphic to Z4 % Z2 % Z2 because Z4 is not isomorphic to Z2 % Z2. To appreciate fully the potency of the Fundamental Theorem, contrast the ease with which the Abelian groups of order pk, k # 4, were determined with the corresponding problem for non-Abelian groups. Even a description of the two non-Abelian groups of order 8 is a challenge (see Chapter 26), and a description of the nine non-Abelian groups of order 16 is well beyond the scope of this text. Now that we know how to construct all the Abelian groups of primepower order, we move to the problem of constructing all Abelian

220

Groups

groups of a certain order n, where n has two or more distinct prime divisors. We begin by writing n in prime-power decomposition form n 5 p1n1p2n2 ? ? ? pk n k. Next, we individually form all Abelian groups of order p1n 1, then p2n 2, and so on, as described earlier. Finally, we form all possible external direct products of these groups. For example, let n 5 1176 5 23 ? 3 ? 72. Then, the complete list of the distinct isomorphism classes of Abelian groups of order 1176 is Z8 % Z3 % Z49, Z4 % Z2 % Z3 % Z49, Z2 % Z2 % Z2 % Z3 % Z49, Z8 % Z3 % Z7 % Z7, Z4 % Z2 % Z3 % Z7 % Z7, Z2 % Z2 % Z2 % Z3 % Z7 % Z7. If we are given any particular Abelian group G of order 1176, the question we want to answer about G is: Which of the preceding six isomorphism classes represents the structure of G? We can answer this question by comparing the orders of the elements of G with the orders of the elements in the six direct products, since it can be shown that two finite Abelian groups are isomorphic if and only if they have the same number of elements of each order. For instance, we could determine whether G has any elements of order 8. If so, then G must be isomorphic to the first or fourth group above, since these are the only ones with elements of order 8. To narrow G down to a single choice, we now need only check whether or not G has an element of order 49, since the first product above has such an element, whereas the fourth one does not. What if we have some specific Abelian group G of order p1n 1p2 n 2 ? ? ? pk n k, where the pi’s are distinct primes? How can G be expressed as an internal direct product of cyclic groups of prime-power order? For simplicity, let us say that the group has 2n elements. First, we must compute the orders of the elements. After this is done, pick an element of maximum order 2r, call it a1. Then a1 is one of the factors in the desired internal direct product. If G 2 a1, choose an element a2 of maximum order 2s such that s # n 2 r and none of a2, a2 2, a2 4, . . . , s21 a22 is in a1. Then a2 is a second direct factor. If n 2 r 1 s, select an element a3 of maximum order 2t such that t # n 2 r 2 s and none of t21 a3, a32, a34, . . . , a32 is in a1 3 a2 5 {a1i a 2 j | 0 # i , 2r, 0 # j , 2s}. Then a3 is another direct factor. We continue in this fashion until our direct product has the same order as G. A formal presentation of this algorithm for any Abelian group G of prime-power order pn is as follows.

221

11 | Fundamental Theorem of Finite Abelian Groups

Greedy Algorithm for an Abelian Group of Order pn 1. Compute the orders of the elements of the group G. 2. Select an element a1 of maximum order and define G1 5 a1. Set i 5 1. 3. If |G| 5 |Gi|, stop. Otherwise, replace i by i 1 1. 4. Select an element ai of maximum order p k such that p k # 2 k21 |G|/|Gi21| and none of ai , ai p, ai p ,. . . , ai p is in Gi21, and define Gi 5 Gi21 3 ai. 5. Return to step 3. In the general case where |G| 5 p1 n 1 p2 n 2 ? ? ? pk n k, we simply use the algorithm to build up a direct product of order p1n 1, then another of order p2 n 2, and so on. The direct product of all of these pieces is the desired factorization of G. The following example is small enough that we can compute the appropriate internal and external direct products by hand. EXAMPLE 1 Let G 5 {1, 8, 12, 14, 18, 21, 27, 31, 34, 38, 44, 47, 51, 53, 57, 64} under multiplication modulo 65. Since G has order 16, we know it is isomorphic to one of Z16, Z8 % Z2, Z4 % Z4, Z4 % Z2 % Z2, Z2 % Z2 % Z2 % Z2. To decide which one, we dirty our hands to calculate the orders of the elements of G. Element 1

8

12

14

18

21

27

31

34 38

44

47 51 53

57 64

1

4

4

2

4

4

4

4

4

4

4

4

Order

4

2

4

2

From the table of orders, we can instantly rule out all but Z4 % Z4 and Z4 % Z2 % Z2 as possibilities. Finally, we observe that since this latter group has a subgroup isomorphic to Z2 % Z2 % Z2, it has more than three elements of order 2, and therefore we must have G L Z4 % Z4. Expressing G as an internal direct product is even easier. Pick an element of maximum order, say the element 8. Then 8 is a factor in the product. Next, choose a second element, say a, so that a has order 4 and a and a2 are not in 8 5 {1, 8, 64, 57}. Since 12 has this property, we have G 5 8 3 12.

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Example 1 illustrates how quickly and easily one can write an Abelian group as a direct product given the orders of the elements of the group. But calculating all those orders is certainly not an appealing prospect! The good news is that, in practice, a combination of theory and calculation of the orders of a few elements will usually suffice. EXAMPLE 2 Let G 5 {1, 8, 17, 19, 26, 28, 37, 44, 46, 53, 62, 64, 71, 73, 82, 89, 91, 98, 107, 109, 116, 118, 127, 134} under multiplication modulo 135. Since G has order 24, it is isomorphic to one of Z8 % Z3 L Z24, Z4 % Z2 % Z3 L Z12 % Z2, Z2 % Z2 % Z2 % Z3 L Z6 % Z2 % Z2. Consider the element 8. Direct calculations show that 86 5 109 and 812 5 1. (Be sure to mod as you go. For example, 83 mod 135 5 512 mod 135 5 107, so compute 84 as 8 ? 107 rather than 8 ? 512.) But now we know G. Why? Clearly, |8| 5 12 rules out the third group in the list. At the same time, |109| 5 2 5 |134| (remember, 134 5 21 mod 135) implies that G is not Z24 (see Theorem 4.4). Thus, G L Z12 % Z2, and G 5 8 3 134. Rather than express an Abelian group as a direct product of cyclic groups of prime-power orders, it is often more convenient to combine the cyclic factors of relatively prime order, as we did in Example 2, to obtain a direct product of the form Zn1 % Zn2 % ? ? ? % Znk , where ni divides ni21. For example, Z4 % Z4 % Z2 % Z9 % Z3 % Z5 would be written as Z180 % Z12 % Z2 (see Exercise 11). The algorithm above is easily adapted to accomplish this by replacing step 4 by 49: select an element ai of maximum order m such that m # |G|/|Gi21| and none of ai, ai 2, . . . , ai m21 is in Gi21, and define Gi 5 Gi21 3 ai. As a consequence of the Fundamental Theorem of Finite Abelian Groups, we have the following corollary, which shows that the converse of Lagrange’s Theorem is true for finite Abelian groups. Corollary Existence of Subgroups of Abelian Groups If m divides the order of a finite Abelian group G, then G has a subgroup of order m.

It is instructive to verify this corollary for a specific case. Let us say that G is an Abelian group of order 72 and we wish to produce a subgroup

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11 | Fundamental Theorem of Finite Abelian Groups

of order 12. According to the Fundamental Theorem, G is isomorphic to one of the following six groups: Z8 % Z9, Z4 % Z2 % Z9, Z2 % Z2 % Z2 % Z9,

Z8 % Z3 % Z3, Z4 % Z2 % Z3 % Z3, Z2 % Z2 % Z2 % Z3 % Z3.

Obviously, Z8 % Z9 L Z72 and Z4 % Z2 % Z3 % Z3 L Z12 % Z6 both have a subgroup of order 12. To construct a subgroup of order 12 in Z4 % Z2 % Z9, we simply piece together all of Z4 and the subgroup of order 3 in Z9; that is, {(a, 0, b) | a [ Z4, b [ {0, 3, 6}}. A subgroup of order 12 in Z8 % Z3 % Z3 is given by {(a, b, 0) | a [ {0, 2, 4, 6}, b [ Z3}. An analogous procedure applies to the remaining cases and indeed to any finite Abelian group.

Proof of the Fundamental Theorem Because of the length and complexity of the proof of the Fundamental Theorem of Finite Abelian Groups, we will break it up into a series of lemmas. Lemma 1 Let G be a finite Abelian group of order pnm, where p is a prime that n does not divide m. Then G 5 H 3 K, where H 5 {x [ G | x p 5 e} and K 5 {x [ G | x m 5 e}. Moreover, |H| 5 pn.

PROOF It is an easy exercise to prove that H and K are subgroups of G (see Exercise 29 in Chapter 3). Because G is Abelian, to prove that G 5 H 3 K we need only prove that G 5 HK and H > K 5 {e}. Since we have gcd(m, pn) 5 1, there are integers s and t such that 1 5 sm 1 tpn. n n For any x in G, we have x 5 x1 5 x sm1tp 5 x smx tp and, by Corollary 4 n of Lagrange’s Theorem (Theorem 7.1), x sm [ H and x tp [ K. Thus, G 5 HK. Now suppose that some x [ H > K. Then x pn 5 e 5 x m and, by Corollary 2 to Theorem 4.1, |x| divides both pn and m. Since p does not divide m, we have | x| 5 1 and, therefore, x 5 e. To prove the second assertion of the lemma, note that pnm 5 |HK| 5 |H||K|/|H > K| 5 |H||K| (see Exercise 7 in the Supplementary Exercises for Chapters 5–8). It follows from Theorem 9.5 and Corollary 2 to Theorem 4.1 that p does not divide |K| and therefore |H| 5 pn. Given an Abelian group G with |G| 5 p1 n 1 p2 n 2 ? ? ? pk n k, where the n p’s are distinct primes, we let G(pi) denote the set {x [ G | x pi 5 e}. i

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It then follows immediately from Lemma 1 and induction that G 5 G( p1) 3 G(p2) 3 ? ? ? 3 G(pk) and |G(pi)| 5 pi n i. Hence, we turn our attention to groups of prime-power order. Lemma 2 Let G be an Abelian group of prime-power order and let a be an element of maximal order in G. Then G can be written in the form a 3 K.

PROOF We denote |G| by pn and induct on n. If n 5 1, then G 5 a 3 e. Now assume that the statement is true for all Abelian groups of order p k, where k , n. Among all the elements of G, choose a of maximal order pm. Then x pm 5 e for all x in G. We may assume that G 2 a, for otherwise there is nothing to prove. Now, among all the elements of G, choose b of smallest order such that b o a. We claim that a > b 5 {e}. Since |b p| 5 |b|/p, we know that b p [ a by the manner in which b was chosen. Say b p 5 ai. Notice that e 5 m21 b pm 5 (b p) pm21 5 (ai) p , so |ai| # p m21. Thus, ai is not a generator of a and, therefore, by Corollary 3 to Theorem 4.2, gcd(p m, i) 2 1. This proves that p divides i, so that we can write i 5 pj. Then b p 5 ai 5 a pj. Consider the element c 5 a2jb. Certainly, c is not in a, for if it were, b would be, too. Also, c p 5 a2jpb p 5 a2ib p 5 b2pb p 5 e. Thus, we have found an element c of order p such that c o a. Since b was chosen to have smallest order such that b o a, we conclude that b also has order p. It now follows that a > b 5 {e} because any nonidentity element of the intersection would generate b and thus contradict b o a. Now consider the factor group G 5 G/b. To simplify the notation, m21 we let x denote the coset xb in G. If |a| , |a| 5 pm, then ap 5 e. This m21 m21 m21 means that (ab) p 5 a p b 5 b, so that ap [ a > b 5 {e}, contradicting the fact that |a| 5 p m. Thus, |a| 5 |a| 5 pm, and therefore a is an element of maximal order in G. By induction, we know that G can be written in the form a 3 K for some subgroup K of G. Let K be the pullback of K under the natural homomorphism from G to G (that is, K 5 {x [ G | x [ K }). We claim that a > K 5 {e}. For if x [ a > K, then x [ a > K 5 {e} 5 b and x [ a > b 5 {e}. It now follows from an order argument (see Exercise 33) that G 5 aK, and therefore G 5 a 3 K. Lemma 2 and induction on the order of the group now give the following.

11 | Fundamental Theorem of Finite Abelian Groups

225

Lemma 3 A finite Abelian group of prime-power order is an internal direct product of cyclic groups.

Let us pause to determine where we are in our effort to prove the Fundamental Theorem of Finite Abelian Groups. The remark following Lemma 1 shows that G 5 G( p1) 3 G( p2) 3 ? ? ? 3 G( pn), where each G( pi) is a group of prime-power order, and Lemma 3 shows that each of these factors is an internal direct product of cyclic groups. Thus, we have proved that G is an internal direct product of cyclic groups of primepower order. All that remains to be proved is the uniqueness of the factors. Certainly the groups G( pi) are uniquely determined by G, since they comprise the elements of G whose orders are powers of pi. So we must prove that there is only one way (up to isomorphism and rearrangement of factors) to write each G( pi) as an internal direct product of cyclic groups. Lemma 4 Suppose that G is a finite Abelian group of prime-power order. If G 5 H1 3 H2 3 ? ? ? 3 Hm and G 5 K1 3 K2 3 ? ? ? 3 Kn, where the H’s and K’s are nontrivial cyclic subgroups with |H1| $ |H2| $ ? ? ? $ |Hm| and |K1| $ |K2| $ ? ? ? $ |Kn|, then m 5 n and |Hi| 5 |Ki| for all i.

PROOF We proceed by induction on |G|. Clearly, the case where |G| 5 p is true. Now suppose that the statement is true for all Abelian groups of order less than |G|. For any Abelian group L, the set L p 5 {x p | x [ L} is a subgroup of L (see Exercise 15 in the Supplementary Exercises for Chapters 1– 4) and, by Theorem 9.5, is a proper subgroup if p divides |L|. It follows that G p 5 H1 p 3 H2 p 3 ? ? ? 3 Hm9 p, and G p 5 K1 p 3 K2 p 3 ? ? ? 3 K n9 ,p where m9 is the largest integer i such that |Hi| . p, and n9 is the largest integer j such that |Kj| . p. (This ensures that our two direct products for G p do not have trivial factors.) Since |G p| , |G|, we have, by induction, m9 5 n9 and |Hi p| 5 |Ki p| for i 5 1, . . . , m9. Since |Hi| 5 p|Hi p|, this proves that |Hi| 5 |Ki| for all i 5 1, . . . , m9. All that remains to be proved is that the number of Hi of order p equals the number of Ki of order p; that is, we must prove that m 2 m9 5 n 2 n9 (since n9 5 m9). This follows directly from the facts that |H1||H2| ? ? ? |Hm9|pm2m9 5 |G| 5 |K1||K2| ? ? ? |Kn9|pn2n9, |Hi| 5 |Ki|, and m9 5 n9.

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Exercises You know it ain’t easy, you know how hard it can be. JOHN LENNON AND PAUL MCCARTNEY,

“The Ballad of John and Yoko”

1. What is the smallest positive integer n such that there are two nonisomorphic groups of order n? Name the two groups. 2. What is the smallest positive integer n such that there are three nonisomorphic Abelian groups of order n? Name the three groups. 3. What is the smallest positive integer n such that there are exactly four nonisomorphic Abelian groups of order n? Name the four groups. 4. Calculate the number of elements of order 2 in each of Z16, Z8 % Z2, Z4 % Z4, and Z4 % Z2 % Z2. Do the same for the elements of order 4. 5. Prove that any Abelian group of order 45 has an element of order 15. Does every Abelian group of order 45 have an element of order 9? 6. Show that there are two Abelian groups of order 108 that have exactly one subgroup of order 3. 7. Show that there are two Abelian groups of order 108 that have exactly four subgroups of order 3. 8. Show that there are two Abelian groups of order 108 that have exactly 13 subgroups of order 3. 9. Suppose that G is an Abelian group of order 120 and that G has exactly three elements of order 2. Determine the isomorphism class of G. 10. Find all Abelian groups (up to isomorphism) of order 360. 11. Prove that every finite Abelian group can be expressed as the (external) direct product of cyclic groups of orders n1, n2, . . . , nt, where ni11 divides ni for i 5 1, 2, . . . , t 2 1. (This exercise is referred to in this chapter and in Chapter 22.) 12. Suppose that the order of some finite Abelian group is divisible by 10. Prove that the group has a cyclic subgroup of order 10. 13. Show, by example, that if the order of a finite Abelian group is divisible by 4, the group need not have a cyclic subgroup of order 4. 14. On the basis of Exercises 12 and 13, draw a general conclusion about the existence of cyclic subgroups of a finite Abelian group.

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227

15. How many Abelian groups (up to isomorphism) are there a. of order 6? b. of order 15? c. of order 42? d. of order pq, where p and q are distinct primes? e. of order pqr, where p, q, and r are distinct primes? f. Generalize parts d and e. 16. How does the number (up to isomorphism) of Abelian groups of order n compare with the number (up to isomorphism) of Abelian groups of order m where a. n 5 32 and m 5 52? b. n 5 24 and m 5 54? c. n 5 pr and m 5 qr, where p and q are prime? d. n 5 pr and m 5 prq, where p and q are distinct primes? e. n 5 pr and m 5 prq2, where p and q are distinct primes? 17. The symmetry group of a nonsquare rectangle is an Abelian group of order 4. Is it isomorphic to Z4 or Z2 % Z2? 18. Verify the corollary to the Fundamental Theorem of Finite Abelian Groups in the case that the group has order 1080 and the divisor is 180. 19. The set {1, 9, 16, 22, 29, 53, 74, 79, 81} is a group under multiplication modulo 91. Determine the isomorphism class of this group. 20. Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic. 21. Characterize those integers n such that the only Abelian groups of order n are cyclic. 22. Characterize those integers n such that any Abelian group of order n belongs to one of exactly four isomorphism classes. 23. Refer to Example 1 in this chapter and explain why it is unnecessary to compute the orders of the last five elements listed to determine the isomorphism class of G. 24. Let G 5 {1, 7, 17, 23, 49, 55, 65, 71} under multiplication modulo 96. Express G as an external and an internal direct product of cyclic groups. 25. Let G 5 {1, 7, 43, 49, 51, 57, 93, 99, 101, 107, 143, 149, 151, 157, 193, 199} under multiplication modulo 200. Express G as an external and an internal direct product of cyclic groups. 26. The set G 5 {1, 4, 11, 14, 16, 19, 26, 29, 31, 34, 41, 44} is a group under multiplication modulo 45. Write G as an external and an internal direct product of cyclic groups of prime-power order.

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27. Suppose that G is an Abelian group of order 9. What is the maximum number of elements (excluding the identity) of which one needs to compute the order to determine the isomorphism class of G? What if G has order 18? What about 16? 28. Suppose that G is an Abelian group of order 16, and in computing the orders of its elements, you come across an element of order 8 and two elements of order 2. Explain why no further computations are needed to determine the isomorphism class of G. 29. Let G be an Abelian group of order 16. Suppose that there are elements a and b in G such that |a| 5 |b| 5 4 and a2 2 b2. Determine the isomorphism class of G. 30. Prove that an Abelian group of order 2n (n $ 1) must have an odd number of elements of order 2. 31. Without using Lagrange’s Theorem, show that an Abelian group of odd order cannot have an element of even order. 32. Let G be the group of all n 3 n diagonal matrices with 61 diagonal entries. What is the isomorphism class of G? 33. Prove the assertion made in the proof of Lemma 2 that G 5 aK. 34. Suppose that G is a finite Abelian group. Prove that G has order pn, where p is prime, if and only if the order of every element of G is a power of p. 35. Dirichlet’s Theorem says that, for every pair of relatively prime integers a and b, there are infinitely many primes of the form at 1 b. Use Dirichlet’s Theorem to prove that every finite Abelian group is isomorphic to a subgroup of a U-group. 36. Determine the isomorphism class of Aut(Z2 % Z3 % Z5). 37. Give an example to show that Lemma 2 is false if G is non-Abelian.

Computer Exercises The purpose of computation is insight, not numbers. RICHARD HAMMING

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software lists the isomorphism classes of all finite Abelian groups of any particular order n. Run the program for n 5 16, 24, 512, 2048, 441000, and 999999.

11 | Fundamental Theorem of Finite Abelian Groups

229

2. This software determines how many integers in a given interval are the order of exactly one Abelian group, of exactly two Abelian groups, and so on, up to exactly nine Abelian groups. Run your program for the integers up to 1000. Then from 10001 to 11000. Then choose your own interval of 1000 consecutive integers. Is there much difference in the results? 3. This software expresses a U-group as an internal direct product of sub-groups H1 3 H2 3 ? ? ? 3 Hi, where |Hi| divides |Hi21|. Run the program for the groups U(32), U(80), and U(65).

Reference 1. R. Hirshon, “On Cancellation in Groups,” American Mathematical Monthly 76 (1969): 1037–1039.

Suggested Readings J. A. Gallian, “Computers in Group Theory,” Mathematics Magazine 49 (1976): 69–73. This paper discusses several computer-related projects in group theory done by undergraduate students. J. Kane, “Distribution of Orders of Abelian Groups,” Mathematics Magazine 49 (1976): 132–135. In this article, the author determines the percentages of integers k between 1 and n, for sufficiently large n, that have exactly one isomorphism class of Abelian groups of order k, exactly two isomorphism classes of Abelian groups of order k, and so on, up to 13 isomorphism classes. G. Mackiw, “Computing in Abstract Algebra,” The College Mathematics Journal 27 (1996): 136–142. This article explains how one can use computer software to implement the algorithm given in this chapter for expressing an Abelian group as an internal direct product.

Suggested Website To find more information about the development of group theory, visit:

http://www-groups.dcs.st-and.ac.uk/~history/

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Supplementary Exercises for Chapters 9–11 Every prospector drills many a dry hole, pulls out his rig, and moves on. JOHN L. HESS

True/false questions for Chapters 9–11 are available on the Web at: http://www.d.umn.edu/,jgallian/TF 1. Suppose that H is a subgroup of G and that each left coset of H in G is some right coset of H in G. Prove that H is normal in G. 2. Use a factor group-induction argument to prove that a finite Abelian group of order n has a subgroup of order m for every positive divisor m of n. 3. Let diag(G) 5 {(g, g) | g [ G}. Prove that diag(G) v G % G if and only if G is Abelian. When G is finite, what is the index of diag(G) in G % G? 4. Let H be any group of rotations in Dn. Prove that H is normal in Dn. 5. Prove that Inn(G) v Aut(G). 6. Let H be a subgroup of G. Prove that H is a normal subgroup if and only if, for all a and b in G, ab [ H implies ba [ H. 7. The factor group GL(2, R)/SL(2, R) is isomorphic to some very familiar group. What is the group? 8. Let k be a divisor of n. The factor group (Z/n)/(k/n) is isomorphic to some very familiar group. What is the group? 9. Let 1 H 5 • £0 0

a 1 0

b c § † a, b, c [ Q ¶ 1

under matrix multiplication. a. Find Z(H) . b. Prove that Z(H) is isomorphic to Q under addition. c. Prove that H/Z(H) is isomorphic to Q % Q. d. Are your proofs for parts a and b valid when Q is replaced by R? Are they valid when Q is replaced by Zp, where p is prime? 10. Prove that D4/Z(D4) is isomorphic to Z2 % Z2. 11. Prove that Q/Z under addition is an infinite group in which every element has finite order. 12. Show that the intersection of any collection of normal subgroups of a group is a normal subgroup.

11 | Supplementary Exercises for Chapters 9–11

231

13. Let n . 1 be a fixed integer and let G be a group. If the set H 5 {x [ G| |x| 5 n} together with the identity forms a subgroup of G, prove that it is a normal subgroup of G. In the case where such a subgroup exists, what can be said about n? Give an example of a non-Abelian group that has such a subgroup. Give an example of a group G and a prime n for which the set H together with the identity is not a subgroup. 14. Show that Q/Z has a unique subgroup of order n for each positive integer n. 15. If H and K are normal Abelian subgroups of a group and H > K 5 {e}, prove that HK is Abelian. 16. Let G be a group of odd order. Prove that the mapping x S x2 from G to itself is one-to-one. 17. Suppose that G is a group of permutations on some set. If |G| 5 60 and orbG(5) 5 {1, 5}, prove that stabG(5) is normal in G. 18. Suppose that G 5 H 3 K and that N is a normal subgroup of H. Prove that N is normal in G. 19. Show that there is no homomorphism from Z8 % Z2 % Z2 onto Z4 % Z4. 20. Show that there is no homomorphism from A4 onto a group of order 2, 4, or 6, but that there is a homomorphism from A4 onto a group of order 3. 21. Let H be a normal subgroup of S4 of order 4. Prove that S4/H is isomorphic to S3. 22. Suppose that f is a homomorphism of U(36), Ker f 5 {1, 13, 25}, and f(5) 5 17. Determine all elements that map to 17. 23. Let n 5 2m, where m is odd. How many elements of order 2 does Dn/Z(Dn) have? How many elements are in the subgroup R360/n/Z(Dn)? How do these numbers compare with the number of elements of order 2 in Dm? 24. Suppose that H is a normal subgroup of a group G of odd order and that |H| 5 5. Show that H # Z(G). 25. Let G be an Abelian group and let n be a positive integer. Let Gn 5 {g | gn 5 e} and Gn 5 {gn | g [ G}. Prove that G/Gn is isomorphic to Gn. 26. Let R1 denote the multiplicative group of positive reals and let T 5 {a 1 bi [ C| a2 1 b2 5 1} be the multiplicative group of complex numbers of norm 1. Show that C* is the internal direct product of R1 and T.

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27. Let G be a finite group and let p be a prime. If p2 . |G|, show that any subgroup of order p is normal in G. 28. Let G 5 Z % Z and H 5 {(x, y)| x and y are even integers}. Show that H is a subgroup of G. Determine the order of G/H. To which familiar group is G/H isomorphic? 29. Let n be a positive integer. Prove that every element of order n in Q/Z is contained in 1/n 1 Z. 30. (1997 Putnam Competition) Let G be a group and let f : G S G be a function such that f(g1)f(g2)f(g3) 5 f(h1)f(h2)f(h3)

31. 32. 33. 34. 35. 36.

37.

38.

whenever g1g2g3 5 e 5 h1h2h3. Prove that there exists an element a in G such that c(x) 5 af(x) is a homomorphism. Prove that every homomorphism from Z % Z into Z has the form (x, y) S ax 1 by, where a and b are integers. Prove that every homomorphism from Z % Z into Z % Z has the form (x, y) S (ax 1 by, cx 1 dy), where a, b, c, and d are integers. Prove that Q/Z is not isomorphic to a proper subgroup of itself. Prove that for each positive integer n, the group Q/Z has exactly f(n) elements of order n (f is the Euler phi function). Show that any group with more than two elements has an automorphism other than the identity mapping. A proper subgroup H of a group G is called maximal if there is no subgroup K such that H , K , G. Prove that Q under addition has no maximal subgroups. Let G be the group of quaternions as given in Exercise 4 of the Supplementary Exercises for Chapters 1–4 and H 5 a2 . Determine whether G>H is isomorphic to Z4 or Z2 % Z2. Is G>H isomorphic to a subgroup of G? Write the dihedral group D8 as 5R0, R45, R90, R135, R180, R225, R270, R315, F1, F2, F3, F4, F5, F6, F7, F8 and let N 5 5R0, R90, R180, R2706 . Prove that N is normal in D8. Given that F1N 5 5F1, F4, F3, F26 determine whether D8>N is cyclic.

11 | Supplementary Exercises for Chapters 9–11

39. Let G be the group e c

1 0

233

a d | where a, b [ R, b ? 0 f and b

1 x d | where x [ R f . Show that H is a subgroup of G. Is 0 1 H a normal subgroup of G? Justify your answer. 40. Find a subgroup H of Zp2 % Zp2 such that (Zp2 % Zp2)/H is isomorphic to Zp % Zp. 41. Recall that H is a characteristic subgroup of K if f(H) 5 H for every automorphism f of K. Prove that if H is a characteristic subgroup of K, and K is a normal subgroup of G, then H is a normal subgroup of G. H5 ec

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PA R T

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Rings

For online student resources, visit this textbook’s website at http://college.hmco.com/PIC/gallian7e

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Introduction 12 to Rings Example is the school of mankind, and they will learn at no other. EDMUND BURKE, On a Regicide Peace

Motivation and Definition Many sets are naturally endowed with two binary operations: addition and multiplication. Examples that quickly come to mind are the integers, the integers modulo n, the real numbers, matrices, and polynomials. When considering these sets as groups, we simply used addition and ignored multiplication. In many instances, however, one wishes to take into account both addition and multiplication. One abstract concept that does this is the concept of a ring.† This notion was originated in the mid-nineteenth century by Richard Dedekind, although its first formal abstract definition was not given until Abraham Fraenkel presented it in 1914. Definition Ring

A ring R is a set with two binary operations, addition (denoted by a 1 b) and multiplication (denoted by ab), such that for all a, b, c in R: 1. a 1 b 5 b 1 a. 2. (a 1 b) 1 c 5 a 1 (b 1 c). 3. There is an additive identity 0. That is, there is an element 0 in R such that a 1 0 5 a for all a in R. 4. There is an element 2a in R such that a 1 (2a) 5 0. 5. a(bc) 5 (ab)c. 6. a(b 1 c) 5 ab 1 ac and (b 1 c) a 5 ba 1 ca.

So, a ring is an Abelian group under addition, also having an associative multiplication that is left and right distributive over addition. Note that multiplication need not be commutative. When it is, we say that the ring is commutative. Also, a ring need not have an identity †The

term ring was first applied in 1897 by the German mathematician David Hilbert (1862–1943). 237

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under multiplication. A unity (or identity) in a ring is a nonzero element that is an identity under multiplication. A nonzero element of a commutative ring with unity need not have a multiplicative inverse. When it does, we say that it is a unit of the ring. Thus, a is a unit if a21 exists. The following terminology and notation are convenient. If a and b belong to a commutative ring R and a is nonzero, we say that a divides b (or that a is a factor of b) and write a | b, if there exists an element c in R such that b 5 ac. If a does not divide b, we write a B b. Recall that if a is an element from a group under the operation of addition and n is a positive integer, na means a 1 a 1 ? ? ? 1 a, where there are n summands. When dealing with rings, this notation can cause confusion, since we also use juxtaposition for the ring multiplication. When there is the potential for confusion, we will use n ? a to mean a 1 a 1 ? ? ? 1 a (n summands). For an abstraction to be worthy of study, it must have many diverse concrete realizations. The following list of examples shows that the ring concept is pervasive.

Examples of Rings EXAMPLE 1 The set Z of integers under ordinary addition and multiplication is a commutative ring with unity 1. The units of Z are 1 and 21. EXAMPLE 2 The set Zn 5 {0, 1, . . . , n 2 1} under addition and multiplication modulo n is a commutative ring with unity 1. The set of units is U(n). EXAMPLE 3 The set Z[x] of all polynomials in the variable x with integer coefficients under ordinary addition and multiplication is a commutative ring with unity f(x) 5 1. EXAMPLE 4 The set M2(Z) of 2 3 2 matrices with integer entries is a noncommutative ring with unity c

1 0 d. 0 1

EXAMPLE 5 The set 2Z of even integers under ordinary addition and multiplication is a commutative ring without unity. EXAMPLE 6 The set of all continuous real-valued functions of a real variable whose graphs pass through the point (1, 0) is a commutative ring without unity under the operations of pointwise addition and

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multiplication [that is, the operations ( f 1 g)(a) 5 f(a) 1 g(a) and ( fg)(a) 5 f(a)g(a)]. EXAMPLE 7 Let R1, R2, . . . , Rn be rings. We can use these to construct a new ring as follows. Let R1 % R2 % ? ? ? % Rn 5 {(a1, a2, . . . , an) | ai [ Ri} and perform componentwise addition and multiplication; that is, define (a1, a2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn) and (a1, a2, . . . , an)(b1, b2, . . . , bn) 5 (a1b1, a2b2, . . . , anbn). This ring is called the direct sum of R1, R2, . . . , Rn.

Properties of Rings Our first theorem shows how the operations of addition and multiplication intertwine. We use b 2 c to denote b 1 (2c). Theorem 12.1 Rules of Multiplication Let a, b, and c belong to a ring R. Then 1. 2. 3. 4.

a0 5 0a 5 0. a(2b) 5 (2a)b 5 2(ab). (2a)(2b) 5 ab.† a(b 2 c) 5 ab 2 ac and (b 2 c)a 5 ba 2 ca.

Furthermore, if R has a unity element 1, then 5. (21)a 5 2a. 6. (21)(21) 5 1.

PROOF We will prove rules 1 and 2 and leave the rest as easy exercises (see Exercise 11). To prove statements such as those in Theorem 12.1, we need only “play off ” the distributive property against the fact that R is a group under addition with additive identity 0. Consider rule 1. Clearly, 0 1 a0 5 a0 5 a(0 1 0) 5 a0 1 a0. So, by cancellation, 0 5 a0. Similarly, 0a 5 0. †Minus

times minus is plus. The reason for this we need not discuss. W. H. Auden

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To prove rule 2, we observe that a(2b) 1 ab 5 a(2b 1 b) 5 a0 5 0. So, adding 2(ab) to both sides yields a(2b) 5 2(ab). The remainder of rule 2 is done analogously. Recall that in the case of groups, the identity and inverses are unique. The same is true for rings, provided that these elements exist. The proofs are identical to the ones given for groups and therefore are omitted. Theorem 12.2 Uniqueness of the Unity and Inverses If a ring has a unity, it is unique. If a ring element has a multiplicative inverse, it is unique.

Many students have the mistaken tendency to treat a ring as if it were a group under multiplication. It is not. The two most common errors are the assumptions that ring elements have multiplicative inverses—they need not—and that a ring has a multiplicative identity—it need not. For example, if a, b, and c belong to a ring, a 2 0 and ab 5 ac, we cannot conclude that b 5 c. Similarly, if a2 5 a, we cannot conclude that a 5 0 or 1 (as is the case with real numbers). In the first place, the ring need not have multiplicative cancellation, and in the second place, the ring need not have a multiplicative identity. There is an important class of rings wherein multiplicative identities exist and for which multiplicative cancellation holds. This class is taken up in the next chapter.

Subrings In our study of groups, subgroups played a crucial role. Subrings, the analogous structures in ring theory, play a much less prominent role than their counterparts in group theory. Nevertheless, subrings are important. Definition Subring

A subset S of a ring R is a subring of R if S is itself a ring with the operations of R.

Just as was the case for subgroups, there is a simple test for subrings. Theorem 12.3 Subring Test A nonempty subset S of a ring R is a subring if S is closed under subtraction and multiplication—that is, if a 2 b and ab are in S whenever a and b are in S.

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PROOF Since addition in R is commutative and S is closed under subtraction, we know by the One-Step Subgroup Test (Theorem 3.1) that S is an Abelian group under addition. Also, since multiplication in R is associative as well as distributive over addition, the same is true for multiplication in S. Thus, the only condition remaining to be checked is that multiplication is a binary operation on S. But this is exactly what closure means. We leave it to the student to confirm that each of the following examples is a subring. EXAMPLE 8 {0} and R are subrings of any ring R. {0} is called the trivial subring of R. EXAMPLE 9 {0, 2, 4} is a subring of the ring Z6, the integers modulo 6. Note that although 1 is the unity in Z6, 4 is the unity in {0, 2, 4}. EXAMPLE 10 For each positive integer n, the set nZ 5 {0, 6n, 62n, 63n, . . .} is a subring of the integers Z. EXAMPLE 11 The set of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z} is a subring of the complex numbers C. EXAMPLE 12 Let R be the ring of all real-valued functions of a single real variable under pointwise addition and multiplication. The subset S of R of functions whose graphs pass through the origin forms a subring of R. EXAMPLE 13 The set ec

a 0 d 0 a, b [ Z f 0 b

of diagonal matrices is a subring of the ring of all 2 3 2 matrices over Z. We can picture the relationship between a ring and its various subrings by way of a subring lattice diagram. In such a diagram, any ring is a subring of all the rings that it is connected to by one or more upward lines. Figure 12.1 shows the relationships among some of the rings we have already discussed.

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Rings

C R Z[i] = {a 1 bi | a, b [ Z}

Q(√2) = {a 1 b√2 | a, b [ Q} Q Z

5Z

10Z

2Z

4Z 8Z

7Z

3Z

6Z 12Z

9Z 18Z

Figure 12.1 Partial subring lattice diagram of C

In the next several chapters, we will see that many of the fundamental concepts of group theory can be naturally extended to rings. In particular, we will introduce ring homomorphisms and factor rings.

Exercises There is no substitute for hard work. THOMAS ALVA EDISON, Life

1. Give an example of a finite noncommutative ring. Give an example of an infinite noncommutative ring that does not have a unity. 2. The ring {0, 2, 4, 6, 8} under addition and multiplication modulo 10 has a unity. Find it. 3. Give an example of a subset of a ring that is a subgroup under addition but not a subring. 4. Show, by example, that for fixed nonzero elements a and b in a ring, the equation ax 5 b can have more than one solution. How does this compare with groups? 5. Prove Theorem 12.2. 6. Find an integer n that shows that the rings Zn need not have the following properties that the ring of integers has. a. a2 5 a implies a 5 0 or a 5 1. b. ab 5 0 implies a 5 0 or b 5 0. c. ab 5 ac and a 2 0 imply b 5 c. Is the n you found prime?

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7. Show that the three properties listed in Exercise 6 are valid for Zp, where p is prime. 8. Show that a ring is commutative if it has the property that ab 5 ca implies b 5 c when a 2 0. 9. Prove that the intersection of any collection of subrings of a ring R is a subring of R. 10. Verify that Examples 8 through 13 in this chapter are as stated. 11. Prove parts 3 through 6 of Theorem 12.1. 12. Let a, b, and c be elements of a commutative ring, and suppose that a is a unit. Prove that b divides c if and only if ab divides c. 13. Describe all the subrings of the ring of integers. 14. Let a and b belong to a ring R and let m be an integer. Prove that m ? (ab) 5 (m ? a)b 5 a(m ? b). 15. Show that if m and n are integers and a and b are elements from a ring, then (m ? a)(n ? b) 5 (mn) ? (ab). (This exercise is referred to in Chapters 13 and 15.) 16. Show that if n is an integer and a is an element from a ring, then n ? (2a) 5 2(n ? a). 17. Show that a ring that is cyclic under addition is commutative. 18. Let a belong to a ring R. Let S 5 {x [ R | ax 5 0}. Show that S is a subring of R. 19. Let R be a ring. The center of R is the set {x [ R | ax 5 xa for all a in R}. Prove that the center of a ring is a subring. 20. Describe the elements of M2(Z) (see Example 4) that have multiplicative inverses. 21. Suppose that R1, R2, . . . , Rn are rings that contain nonzero elements. Show that R1 % R2 % ? ? ? % Rn has a unity if and only if each Ri has a unity. 22. Let R be a commutative ring with unity and let U(R) denote the set of units of R. Prove that U(R) is a group under the multiplication of R. (This group is called the group of units of R.) 23. Determine U(Z[i]) (see Example 11). 24. If R1, R2, . . . , Rn are commutative rings with unity, show that U(R1 % R2 % ? ? ? % Rn) 5 U(R1) % U(R2) % ? ? ? % U(Rn). 25. Determine U(Z[x]). (This exercise is referred to in Chapter 17.) 26. Determine U(R[x]). 27. Show that a unit of a ring divides every element of the ring. 28. In Z6, show that 4 | 2; in Z8, show that 3 | 7; in Z15, show that 9 | 12.

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29. Suppose that a and b belong to a commutative ring R with unity. If a is a unit of R and b2 5 0, show that a 1 b is a unit of R. 30. Suppose that there is an integer n . 1 such that xn 5 x for all elements x of some ring. If m is a positive integer and am 5 0 for some a, show that a 5 0. 31. Give an example of ring elements a and b with the properties that ab 5 0 but ba 2 0. 32. Let n be an integer greater than 1. In a ring in which xn 5 x for all x, show that ab 5 0 implies ba 5 0. 33. Suppose that R is a ring such that x3 5 x for all x in R. Prove that 6x 5 0 for all x in R. 34. Suppose that a belongs to a ring and a4 5 a2. Prove that a2n 5 a2 for all n $ 1. 35. Find an integer n . 1 such that an 5 a for all a in Z6. Do the same for Z10. Show that no such n exists for Zm when m is divisible by the square of some prime. 36. Let m and n be positive integers and let k be the least common multiple of m and n. Show that mZ > nZ 5 kZ. 37. Explain why every subgroup of Zn under addition is also a subring of Zn. 38. Is Z6 a subring of Z12? 39. Suppose that R is a ring with unity 1 and a is an element of R such that a2 5 1. Let S 5 {ara | r [ R}. Prove that S is a subring of R. Does S contain 1? 40. Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5 a a1b ec d ` a, b [ Z f . Prove or disprove that R is a subring a1b b 41.

42. 43. 44.

of M2(Z). Let M2(Z) be the ring of all 2 3 2 matrices over the integers and let R 5 a a2b ec d ` a, b [ Z f . Prove or disprove that R is a suba2b b ring of M2(Z). a a Let R 5 e c d ` a, b [ Z f . Prove or disprove that R is a subring b b of M2(Z). Let R 5 Z % Z % Z and S 5 {(a, b, c) [ R | a 1 b 5 c}. Prove or disprove that S is a subring of R. Suppose that there is a positive even integer n such that an 5 a for all elements a of some ring. Show that 2a 5 a for all a in the ring.

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45. Let R be a ring with unity 1. Show that S 5 {n ? 1| n [ Z} is a subring of R. 46. Show that 2Z < 3Z is not a subring of Z. 47. Determine the smallest subring of Q that contains 1/2. (That is, find the subring S with the property that S contains 1/2 and, if T is any subring containing 1/2, then T contains S.) 48. Determine the smallest subring of Q that contains 2/3. 49. Let R be a ring. Prove that a2 2 b2 5 (a 1 b)(a 2 b) for all a, b in R if and only if R is commutative. 50. Suppose that R is a ring and that a2 5 a for all a in R. Show that R is commutative. [A ring in which a2 5 a for all a is called a Boolean ring, in honor of the English mathematician George Boole (1815–1864).] 51. Give an example of a Boolean ring with four elements. Give an example of an infinite Boolean ring.

Computer Exercises Theory is the general; experiments are the soldiers. LEONARDO DA VINCI

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software finds all solutions to the equation x2 1 y2 5 0 in Zp. Run the software for all odd primes up to 37. Make a conjecture about the existence of nontrivial solutions in Zp ( p a prime) and the form of p. 2. Let Zn[i] 5 {a 1 bi| a, b belong to Zn, i2 5 21} (the Gaussian integers modulo n). This software finds the group of units of this ring and the order of each element of the group. Run the program for n 5 3, 7, 11, and 23. Is the group of units cyclic for these cases? Try to guess a formula for the order of the group of units of Zn[i] as a function of n when n is a prime and n mod 4 5 3. Run the program for n 5 9 and 27. Are the groups cyclic? Try to guess a formula for the order when n 5 3k. Run the program for n 5 5, 13, 17, and 29. Is the group cyclic for these cases? What is the largest order of any element in the group? Try to guess a formula for the order of the group of units of Zn[i] as a function of n when n is a prime and n mod 4 5 1. Try to guess a formula for the largest order of any

246

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element in the group of units of Zn[i] as a function of n when n is a prime and n mod 4 5 1. On the basis of the orders of the elements of the group of units, try to guess the isomorphism class of the group. Run the program for n 5 25. Is this group cyclic? Based on the number of elements in this group and the orders of the elements, try to guess the isomorphism class of the group. 3. This software determines the isomorphism class of the group of units of Zn[i]. Run the program for n 5 5, 13, 17, 29, and 37. Make a conjecture. Run the program for n 5 3, 7, 11, 19, 23, and 31. Make a conjecture. Run the program for n 5 5, 25, and 125. Make a conjecture. Run the program for n 5 13 and 169. Make a conjecture. Run the program for n 5 3, 9, and 27. Make a conjecture. Run the program for n 5 7 and 49. Make a conjecture. Run the program for n 5 11 and 121. Make a conjecture. Make a conjecture about the case where n 5 pk where p is a prime and p mod 4 5 1. Make a conjecture about the case where n 5 pk where p is a prime and p mod 4 5 3. 4. This software determines the order of the group of units in the ring of 2 3 2 matrices over Zn (that is, the group GL(2, Zn)) and the subgroup SL(2, Zn). Run the program for n 5 2, 3, 5, 7, 11, and 13. What relationship do you see between the order of GL(2, Zn) and the order of SL(2, Zn) in these cases? Run the program for n 516, 27, 25, and 49. Make a conjecture about the relationship between the order of GL(2, Zn) and the order of SL(2, Zn) when n is a power of a prime. Run the program for n 5 32. (Notice that when you run the program for n 5 32, the table shows the orders for all divisors of 32 greater than 1.) How do the orders of the two groups change each time you increase the power of 2 by 1? Run the program for n 5 27. How do the orders of the two groups change each time you increase the power of 3 by 1? Run the program for n 5 25. How do the orders of the two groups change when you increase the power of 5 by 1? Make a conjecture about the relationship between |SL(2, Zp i)| and |SL(2, Zp i11)|. Make a conjecture about the relationship between |GL(2, Zpt)| and |GL(2, Zp i11)|. Run the program for n 5 12, 15, 20, 21, and 30. Make a conjecture about the order of GL(2, Zn) in terms of the orders of GL(2, Zs) and GL(2, Zt) where n 5 st and s and t are relatively prime. (Notice that when you run the program for st, the table shows the values for st, s, and t.) For each value of n, is the order of SL(2, Zn) divisible by n? Is it divisible by n 1 1? Is it divisible by n 2 1? 5. In the ring Zn, this software finds the number of solutions to the equation x2 5 21. Run the program for all primes between 3 and 29.

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How does the answer depend on the prime? Make a conjecture about the number of solutions when n is a prime greater than 2. Run the program for the squares of all primes between 3 and 29. Make a conjecture about the number of solutions when n is the square of a prime greater than 2. Run the program for the cubes of primes between 3 and 29. Make a conjecture about the number of solutions when n is any power of an odd prime. Run the program for n 5 2, 4, 8, 16, and 32. Make a conjecture about the number of solutions when n is a power of 2. Run the program for n 5 12, 20, 24, 28, and 36. Make a conjecture about the number of solutions when n is a multiple of 4. Run the program for various cases where n 5 pq and n 5 2pq where p and q are odd primes. Make a conjecture about the number of solutions when n 5 pq or n 5 2pq where p and q are odd primes. What relationship do you see among the numbers of solutions for n 5 p, n 5 q, and n 5 pq? Run the program for various cases where n 5 pqr and n 5 2pqr where p, q, and r are odd primes. Make a conjecture about the number of solutions when n 5 pqr or n 5 2pqr where p, q, and r are odd primes. What relationship do you see among the numbers of solutions when n 5 p, n 5 q, and n 5 r and the case that n 5 pqr? 6. This software determines the number of solutions to the equation X2 5 2I where X is a 2 3 2 matrix with entries from Zn and I is the identity. Run the program for n 5 32. Make a conjecture about the number of solutions when n 5 2k where k . 1. Run the program for n 5 3, 11, 19, 23, and 31. Make a conjecture about the number of solutions when n is a prime of the form 4q 1 3. Run the program for n 5 27 and 49. Make a conjecture about the number of solutions when n has the form pi where p is a prime of the form 4q 1 3. Run the program for n 5 5, 13, 17, 29, and 37. Make a conjecture about the number of solutions when n is a prime of the form 4q 1 1. Run the program for n 5 6, 10, 14, 22, 15, 21, 33, 39, 30, 42. What seems to be the relationship between the number of solutions for a given n and the number of solutions for the prime power factors of n?

Suggested Reading D. B. Erickson, “Orders for Finite Noncommutative Rings,” American Mathematical Monthly 73 (1966): 376–377. In this elementary paper, it is shown that there exists a noncommutative ring of order m . 1 if and only if m is divisible by the square of a prime.

I. N. Herstein A whole generation of textbooks and an entire generation of mathematicians, myself included, have been profoundly influenced by that text [Herstein’s Topics in Algebra]. GEORGIA BENKART

I. N. HERSTEIN was born on March 28, 1923, in Poland. His family moved to Canada when he was seven. He grew up in a poor and tough environment, on which he commented that in his neighborhood you became either a gangster or a college professor. During his school years he played football, hockey, golf, tennis, and pool. During this time he worked as a steeplejack and as a barber at a fair. Herstein received a B.S. degree from the University of Manitoba, an M.A. from the University of Toronto, and, in 1948, a Ph.D. degree from Indiana University under the supervision of Max Zorn. Before permanently settling at the University of Chicago in 1962, he held positions at the University of Kansas, the Ohio State University, the University of Pennsylvania, and Cornell University. Herstein wrote more than 100 research papers and a dozen books. Although his principal interest was noncommutative ring

248

theory, he also wrote papers on finite groups, linear algebra, and mathematical economics. His textbook Topics in Algebra, first published in 1964, dominated the field for 20 years and has become a classic. Herstein had great influence through his teaching and his collaboration with colleagues. He had 30 Ph.D. students, and traveled and lectured widely. His nonmathematical interests included languages and art. He spoke Italian, Hebrew, Polish, and Portuguese. Herstein died on February 9, 1988, after a long battle with cancer. To find more information about Herstein, visit: http://www-groups.dcs.stand.ac.uk/~history/

13 Integral Domains

Don’t just read it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? Where does the proof use the hypothesis? PAUL HALMOS

Definition and Examples To a certain degree, the notion of a ring was invented in an attempt to put the algebraic properties of the integers into an abstract setting. A ring is not the appropriate abstraction of the integers, however, for too much is lost in the process. Besides the two obvious properties of commutativity and existence of a unity, there is one other essential feature of the integers that rings in general do not enjoy—the cancellation property. In this chapter, we introduce integral domains—a particular class of rings that have all three of these properties. Integral domains play a prominent role in number theory and algebraic geometry. Definition Zero-Divisors

A zero-divisor is a nonzero element a of a commutative ring R such that there is a nonzero element b [ R with ab 5 0. Definition Integral Domain

An integral domain is a commutative ring with unity and no zero-divisors.

Thus, in an integral domain, a product is 0 only when one of the factors is 0; that is, ab 5 0 only when a 5 0 or b 5 0. The following examples show that many familiar rings are integral domains and some familiar rings are not. For each example, the student should verify the assertion made. EXAMPLE 1 The ring of integers is an integral domain.

249

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Rings

EXAMPLE 2 The ring of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z} is an integral domain. EXAMPLE 3 The ring Z[x] of polynomials with integer coefficients is an integral domain. EXAMPLE 4 The ring Z["2 ] 5 {a 1 b"2 | a, b [ Z} is an integral domain. EXAMPLE 5 The ring Zp of integers modulo a prime p is an integral domain. EXAMPLE 6 The ring Zn of integers modulo n is not an integral domain when n is not prime. EXAMPLE 7 The ring M2(Z) of 2 3 2 matrices over the integers is not an integral domain. EXAMPLE 8 Z % Z is not an integral domain. What makes integral domains particularly appealing is that they have an important multiplicative group-theoretic property, in spite of the fact that the nonzero elements need not form a group under multiplication. This property is cancellation. Theorem 13.1 Cancellation Let a, b, and c belong to an integral domain. If a 2 0 and ab 5 ac, then b 5 c.

PROOF From ab 5 ac, we have a(b 2 c) 5 0. Since a 2 0, we must have b 2 c 5 0. Many authors prefer to define integral domains by the cancellation property—that is, as commutative rings with unity in which the cancellation property holds. This definition is equivalent to ours.

Fields In many applications, a particular kind of integral domain called a field is necessary. Definition Field

A field is a commutative ring with unity in which every nonzero element is a unit.

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251

To verify that every field is an integral domain, observe that if a and b belong to a field with a 2 0 and ab 5 0, we can multiply both sides of the last expression by a21 to obtain b 5 0. It is often helpful to think of ab21 as a divided by b. With this in mind, a field can be thought of as simply an algebraic system that is closed under addition, subtraction, multiplication, and division (except by 0). We have had numerous examples of fields: the complex numbers, the real numbers, the rational numbers. The abstract theory of fields was initiated by Heinrich Weber in 1893. Groups, rings, and fields are the three main branches of abstract algebra. Theorem 13.2 says that, in the finite case, fields and integral domains are the same. Theorem 13.2 Finite Integral Domains Are Fields A finite integral domain is a field.

PROOF Let D be a finite integral domain with unity 1. Let a be any nonzero element of D. We must show that a is a unit. If a 5 1, a is its own inverse, so we may assume that a 2 1. Now consider the following sequence of elements of D: a, a2, a3, . . . . Since D is finite, there must be two positive integers i and j such that i . j and ai 5 a j. Then, by cancellation, ai2j 5 1. Since a 2 1, we know that i 2 j . 1, and we have shown that ai2j21 is the inverse of a. Corollary Zp Is a Field For every prime p, Zp, the ring of integers modulo p, is a field.

PROOF According to Theorem 13.2, we need only prove that Zp has no zero-divisors. So, suppose that a, b [ Zp and ab 5 0. Then ab 5 pk for some integer k. But then, by Euclid’s Lemma (see Chapter 0), p divides a or p divides b. Thus, in Zp, a 5 0 or b 5 0. Putting the preceding corollary together with Example 6, we see that Zn is a field if and only if n is prime. In Chapter 22, we will describe how all finite fields can be constructed. For now, we give one example of a finite field that is not of the form Zp. EXAMPLE 9 Field with Nine Elements Let Z3[i] 5 {a 1 bi | a, b [ Z3} 5 {0, 1, 2, i, 1 1 i, 2 1 i, 2i, 1 1 2i, 2 1 2i},

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where i2 5 21. This is the ring of Gaussian integers modulo 3. Elements are added and multiplied as in the complex numbers, except that the coefficients are reduced modulo 3. In particular, 21 5 2. Table 13.1 is the multiplication table for the nonzero elements of Z3[i]. Table 13.1 Multiplication Table for Z3[i]*

1 2 i 11i 21i 2i 1 1 2i 2 1 2i

1

2

i

11i

21i

2i

1 1 2i

2 1 2i

1 2 i 11i 21i 2i 1 1 2i 2 1 2i

2 1 2i 2 1 2i 1 1 2i i 21i 11i

i 2i 2 21i 2 1 2i 1 11i 1 1 2i

11i 2 1 2i 21i 2i 1 1 1 2i 2 i

21i 1 1 2i 2 1 2i 1 i 11i 2i 2

2i i 1 1 1 2i 11i 2 2 1 2i 21i

1 1 2i 21i 11i 2 2i 2 1 2i i 1

2 1 2i 11i 1 1 2i i 2 21i 1 2i

EXAMPLE 10 Let Q["2] 5 {a 1 b"2 | a, b [ Q}. It is easy to see that Q["2] is a ring. Viewed as an element of R, the multiplicative inverse of any nonzero element of the form a 1 b"2 is simply 1/(a 1 b"2). To verify that Q["2] is a field, we must show that 1/(a 1 b"2) can be written in the form c 1 d"2. In high school algebra, this process is called “rationalizing the denominator.” Specifically, 1 1 a 2 b"2 a b 5 5 2 "2. 2 2 2 a 2 2b a 2 2b2 a 1 b"2 a 1 b"2 a 2 b"2

(Note that a 1 b"2 2 0 guarantees that a 2 b"2 2 0.)

Characteristic of a Ring Note that for any element x in Z3[i], we have 3x 5 x 1 x 1 x 5 0, since addition is done modulo 3. Similarly, in the subring {0, 3, 6, 9} of Z12, we have 4x 5 x 1 x 1 x 1 x 5 0 for all x. This observation motivates the following definition. Definition Characteristic of a Ring

The characteristic of a ring R is the least positive integer n such that nx 5 0 for all x in R. If no such integer exists, we say that R has characteristic 0. The characteristic of R is denoted by char R.

Thus, the ring of integers has characteristic 0, and Zn has characteristic n. An infinite ring can have a nonzero characteristic. Indeed, the

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ring Z2[x] of all polynomials with coefficients in Z2 has characteristic 2. (Addition and multiplication are done as for polynomials with ordinary integer coefficients except that the coefficients are reduced modulo 2.) When a ring has a unity, the task of determining the characteristic is simplified by Theorem 13.3. Theorem 13.3 Characteristic of a Ring with Unity Let R be a ring with unity 1. If 1 has infinite order under addition, then the characteristic of R is 0. If 1 has order n under addition, then the characteristic of R is n.

PROOF If 1 has infinite order, then there is no positive integer n such that n ? 1 5 0, so R has characteristic 0. Now suppose that 1 has additive order n. Then n ? 1 5 0, and n is the least positive integer with this property. So, for any x in R, we have n ? x 5 x 1 x 1 ? ? ? 1 x (n summands) 5 1x 1 1x 1 ? ? ? 1 1x (n summands) 5 (1 1 1 1 ? ? ? 1 1)x (n summands) 5 (n ? 1)x 5 0x 5 0. Thus, R has characteristic n. In the case of an integral domain, the possibilities for the characteristic are severely limited. Theorem 13.4 Characteristic of an Integral Domain The characteristic of an integral domain is 0 or prime.

PROOF By Theorem 13.3, it suffices to show that if the additive order of 1 is finite, it must be prime. Suppose that 1 has order n and that n 5 st, where 1 # s, t # n. Then, by Exercise 15 in Chapter 12, 0 5 n ? 1 5 (st) ? 1 5 (s ? 1)(t ? 1). So, s ? 1 5 0 or t ? 1 5 0. Since n is the least positive integer with the property that n ? 1 5 0, we must have s 5 n or t 5 n. Thus, n is prime. We conclude this chapter with a brief discussion of polynomials with coefficients from a ring—a topic we will consider in detail in

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later chapters. The existence of zero-divisors in a ring causes unusual results when one is finding zeros of polynomials with coefficients in the ring. Consider, for example, the equation x2 2 4x 1 3 5 0. In the integers, we could find all solutions by factoring x2 2 4x 1 3 5 (x 2 3)(x 2 1) 5 0 and setting each factor equal to 0. But notice that when we say we can find all solutions in this manner, we are using the fact that the only way for a product to equal 0 is for one of the factors to be 0—that is, we are using the fact that Z is an integral domain. In Z12, there are many pairs of nonzero elements whose products are 0: 2 ? 6 5 0, 3 ? 4 5 0, 4 ? 6 5 0, 6 ? 8 5 0, and so on. So, how do we find all solutions of x2 2 4x 1 3 5 0 in Z12? The easiest way is simply to try every element! Upon doing so, we find four solutions: x 5 1, x 5 3, x 5 7, and x 5 9. Observe that we can find all solutions of x2 2 4x 1 3 5 0 over Z11 or Z13, say, by setting the two factors x 2 3 and x 2 1 equal to 0. Of course, the reason this works for these rings is that they are integral domains. Perhaps this will convince you that integral domains are particularly advantageous rings. Table 13.2 gives a summary of some of the rings we have introduced and their properties. Table 13.2 Summary of Rings and Their Properties Ring Z Zn, n composite Zp, p prime Z[x] nZ, n . 1 M2(Z) M2(2Z) Z[i] Z3[i] Z["2] Q["2] Z%Z

Form of Element

Unity

Integral Commutative Domain Field Characteristic

k k k anxn 1 ? ? ? 1 a1x 1 a0 nk

1 1 1 f(x) 5 1

Yes Yes Yes Yes

Yes No Yes Yes

No No Yes No

0 n p 0

None

Yes

No

No

0

a b c d c d 2a 2b c d 2c 2d

1 0 c d 0 1

No

No

No

0

None

No

No

No

0

1 1 1 1 (1, 1)

Yes Yes Yes Yes Yes

Yes Yes Yes Yes No

No Yes No Yes No

0 3 0 0 0

a 1 bi a 1 bi; a, b [ Z3 a 1 b"2; a, b [ Z a 1 b"2; a, b [ Q (a, b)

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Exercises It looked absolutely impossible. But it so happens that you go on worrying away at a problem in science and it seems to get tired, and lies down and lets you catch it. WILLIAM LAWRENCE BRAGG†

1. Verify that Examples 1 through 8 are as claimed. 2. Which of Examples 1 through 5 are fields? 3. Show that a commutative ring with the cancellation property (under multiplication) has no zero-divisors. 4. List all zero-divisors in Z20. Can you see a relationship between the zero-divisors of Z20 and the units of Z20? 5. Show that every nonzero element of Zn is a unit or a zero-divisor. 6. Find a nonzero element in a ring that is neither a zero-divisor nor a unit. 7. Let R be a finite commutative ring with unity. Prove that every nonzero element of R is either a zero-divisor or a unit. What happens if we drop the “finite” condition on R? 8. Describe all zero-divisors and units of Z % Q % Z. 9. Let d be an integer. Prove that Z["d ] 5 {a 1 b"d | a, b [ Z} is an integral domain. (This exercise is referred to in Chapter 18.) 10. In Z7, give a reasonable interpretation for the expressions 1/2, 22/3, "23, and 21/6. 11. Give an example of a commutative ring without zero-divisors that is not an integral domain. 12. Find two elements a and b in a ring such that both a and b are zerodivisors, a 1 b 2 0, and a 1 b is not a zero-divisor. 13. Let a belong to a ring R with unity and suppose that an 5 0 for some positive integer n. (Such an element is called nilpotent.) Prove that 1 2 a has a multiplicative inverse in R. [Hint: Consider (1 2 a)(1 1 a 1 a2 1 ? ? ? 1 an21).] 14. Show that the nilpotent elements of a commutative ring form a subring. 15. Show that 0 is the only nilpotent element in an integral domain. 16. A ring element a is called an idempotent if a2 5 a. Prove that the only idempotents in an integral domain are 0 and 1. †Bragg, at age 24, won the Nobel Prize for the invention of x-ray crystallography. He remains the youngest person ever to receive the Nobel Prize.

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17. Let a and b be idempotents in a commutative ring. Show that each of the following is also an idempotent: ab, a 2 ab, a 1 b 2 ab, a 1 b 2 2ab. 18. Prove that if a is a ring idempotent, then an 5 a for all positive integers n. 19. Determine all ring elements that are both nilpotents and idempotents. 20. Find a zero-divisor in Z5[i] 5 {a 1 bi | a, b [ Z5}. 21. Find an idempotent in Z5[i] 5 {a 1 bi | a, b [ Z5}. 22. Find all units, zero-divisors, idempotents, and nilpotent elements in Z3 % Z6. 23. Determine all elements of a ring that are both units and idempotents. 24. Let R be the set of all real-valued functions defined for all real numbers under function addition and multiplication. a. Determine all zero-divisors of R. b. Determine all nilpotent elements of R. c. Show that every nonzero element is a zero-divisor or a unit. 25. (Subfield Test) Let F be a field and let K be a subset of F with at least two elements. Prove that K is a subfield of F if, for any a, b (b 2 0) in K, a 2 b and ab21 belong to K. 26. Let d be a positive integer. Prove that Q["d ] 5 {a 1 b"d | a, b [ Q} is a field. 27. Let R be a ring with unity 1. If the product of any pair of nonzero elements of R is nonzero, prove that ab 5 1 implies ba 5 1. 28. Let R 5 {0, 2, 4, 6, 8} under addition and multiplication modulo 10. Prove that R is a field. 29. Formulate the appropriate definition of a subdomain (that is, a “sub” integral domain). Let D be an integral domain with unity 1. Show that P 5 {n ? 1 | n [ Z} (that is, all integral multiples of 1) is a subdomain of D. Show that P is contained in every subdomain of D. What can we say about the order of P? 30. Prove that there is no integral domain with exactly six elements. Can your argument be adapted to show that there is no integral domain with exactly four elements? What about 15 elements? Use these observations to guess a general result about the number of elements in a finite integral domain. 31. Let F be a field of order 2n. Prove that char F 5 2. 32. Determine all elements of an integral domain that are their own inverses under multiplication. 33. Characterize those integral domains for which 1 is the only element that is its own multiplicative inverse.

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34. Determine all integers n . 1 for which (n 2 1) ! is a zero-divisor in Zn. 35. Suppose that a and b belong to an integral domain. a. If a5 5 b5 and a3 5 b3, prove that a 5 b. b. If am 5 bm and an 5 bn, where m and n are positive integers that are relatively prime, prove that a 5 b. 36. Find an example of an integral domain and distinct positive integers m and n such that am 5 bm and an 5 bn, but a 2 b. 37. If a is an idempotent in a commutative ring, show that 1 2 a is also an idempotent. 38. Construct a multiplication table for Z2[i], the ring of Gaussian integers modulo 2. Is this ring a field? Is it an integral domain? 39. The nonzero elements of Z3[i] form an Abelian group of order 8 under multiplication. Is it isomorphic to Z8, Z4 % Z2, or Z2 % Z2 % Z2? 40. Show that Z 7["3] 5 {a 1 b"3 | a, b [ Z 7} is a field. For any positive integer k and any prime p, determine a necessary and sufficient condition for Zp["k] 5 {a 1 b"k | a, b [ Zp} to be a field. 41. Show that a finite commutative ring with no zero-divisors and at least two elements has a unity. 42. Suppose that a and b belong to a commutative ring and ab is a zero-divisor. Show that either a or b is a zero-divisor. 43. Suppose that R is a commutative ring without zero-divisors. Show that all the nonzero elements of R have the same additive order. 44. Suppose that R is a commutative ring without zero-divisors. Show that the characteristic of R is 0 or prime. 45. Let x and y belong to a commutative ring R with prime characteristic p. a. Show that (x 1 y) p 5 x p 1 y p. n n n b. Show that, for all positive integers n, (x 1 y) p 5 x p 1 y p . c. Find elements x and y in a ring of characteristic 4 such that (x 1 y)4 2 x4 1 y4. (This exercise is referred to in Chapter 20.) 46. Let R be a commutative ring with unity 1 and prime characteristic. If a [ R is nilpotent, prove that there is a positive integer k such that (1 1 a)k 5 1. 47. Show that any finite field has order pn, where p is a prime. Hint: Use facts about finite Abelian groups. (This exercise is referred to in Chapter 22.) 48. Give an example of an infinite integral domain that has characteristic 3. 49. Let R be a ring and let M2(R) be the ring of 2 3 2 matrices with entries from R. Explain why these two rings have the same characteristic.

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50. Let R be a ring with m elements. Show that the characteristic of R divides m. 51. Explain why a finite ring must have a nonzero characteristic. 52. Find all solutions of x2 2 x 1 2 5 0 over Z3[i]. (See Example 9.) 53. Consider the equation x2 2 5x 1 6 5 0. a. How many solutions does this equation have in Z7? b. Find all solutions of this equation in Z8. c. Find all solutions of this equation in Z12. d. Find all solutions of this equation in Z14. 54. Find the characteristic of Z4 % 4Z. 55. Suppose that R is an integral domain in which 20 ? 1 5 0 and 12 ? 1 5 0. (Recall that n ? 1 means the sum 1 1 1 1 ? ? ? 1 1 with n terms.) What is the characteristic of R? 56. In a commutative ring of characteristic 2, prove that the idempotents form a subring. 57. Describe the smallest subfield of the field of real numbers that contains "2. (That is, describe the subfield K with the property that K contains "2 and if F is any subfield containing "2, then F contains K.) 58. Let F be a finite field with n elements. Prove that xn21 5 1 for all nonzero x in F. 59. Let F be a field of prime characteristic p. Prove that K 5 {x [ F | x p 5 x} is a subfield of F. 60. Suppose that a and b belong to a field of order 8 and that a2 1 ab 1 b2 5 0. Prove that a 5 0 and b 5 0. Do the same when the field has order 2n with n odd. 61. Let F be a field of characteristic 2 with more than two elements. Show that (x 1 y)3 2 x3 1 y3 for some x and y in F. 62. Suppose that F is a field with characteristic not 2, and that the nonzero elements of F form a cyclic group under multiplication. Prove that F is finite. 63. Suppose that D is an integral domain and that f is a nonconstant function from D to the nonnegative integers such that f(xy) 5 f(x)f(y). If x is a unit in D, show that f(x) 5 1. 64. Let F be a field of order 32. Show that the only subfields of F are F itself and {0, 1}. 65. Suppose that F is a field with 27 elements. Show that for every element a [ F , 5a 5 2a.

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Computer Exercises The basic unit of mathematics is conjecture. ARNOLD ROSS

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software lists the idempotents (see Exercise 16 for the definition) in Zn. Run the program for various values of n. Use these data to make conjectures about the number of idempotents in Zn as a function of n. For example, how many idempotents are there when n is a prime power? What about when n is divisible by exactly two distinct primes? In the case where n is of the form pq where p and q are primes, can you see a relationship between the two idempotents that are not 0 and 1? Can you see a relationship between the number of idempotents for a given n and the number of distinct prime divisors of n? 2. This software lists the nilpotent elements (see Exercise 13 for the definition) in Zn. Run the program for various values of n. Use these data to make conjectures about the number of nilpotent elements in Zn as a function of n. 3. This software determines which rings of the form Zp[i] are fields. Run the program for all primes up to 37. From these data, make a conjecture about the form of the primes that yield a field. 4. This software finds the idempotents in Zn[i] 5 {a 1 bi | a, b [ Zn} (Gaussian integers modulo n). Run the software for n 5 4, 8, 16, and 32. Make a conjecture about the number of idempotents when n 5 2k. Run the software for n 5 13, 17, 29, and 37. What do these values of n have in common? Make a conjecture about the number of idempotents for these n. Run the software for n 5 7, 11, 19, 23, 31, and 43. What do these values of n have in common? Make a conjecture about the number of idempotents for these n. 5. This software finds the nilpotent elements in Zn[i] 5 {a 1 bi | a, b [ Zn}. Run the software for n 5 4, 8, 16, and 32. Make a conjecture about the number of nilpotent elements when n 5 2k. Run the software for n 5 3, 5, 7, 11, 13, and 17. What do these values of n have in common? Make a conjecture about the number of nilpotent elements for these n. Run the program for n 5 9. Do you need to revise the conjecture you made based on n 5 3, 5, 7, 11, 13, and 17? Run the software for n 5 9, 25, and 49. What do these values

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of n have in common? Make a conjecture about the number of nilpotent elements for these n. Run the program for n 5 81. Do you need to revise the conjecture you made based on n 59, 25, and 49? What do these values of n have in common? Make a conjecture about the number of nilpotent elements for these n. Run the program for n 5 27. Do you need to revise the conjecture you made based on n 5 9, 25, and 49? Run your program for n 5 125 (this may take a few seconds). On the basis of all of your data for this exercise, make a single conjecture in the case that n 5 pk where p is any prime. Run the program for n 5 6, 15, and 21. Make a conjecture. Run the program for 12, 20, 28, and 45. Make a conjecture. Run the program for 36 and 100 (this may take a few minutes). On the basis of all your data for this exercise, make a single conjecture that covers all integers n . 1. 6. This software determines the zero-divisors in Zn[i] 5 {a 1 bi | a, b [ Zn}. Use the software to formulate and test conjectures about the number of zero-divisors in Zn[i] based on various conditions of n.

Suggested Readings Eric Berg, “A Family of Fields,” Pi Mu Epsilon 9 (1990): 154–155. In this article, the author uses properties of logarithms and exponents to define recursively an infinite family of fields starting with the real numbers. N. A. Khan, “The Characteristic of a Ring,” American Mathematical Monthly 70 (1963): 736. Here it is shown that a ring has nonzero characteristic n if and only if n is the maximum of the orders of the elements of R. K. Robin McLean, “Groups in Modular Arithmetic,” The Mathematical Gazette 62 (1978): 94–104. This article explores the interplay between various groups of integers under multiplication modulo n and the ring Zn. It shows how to construct groups of integers in which the identity is not obvious; for example, 1977 is the identity of the group {1977, 5931} under multiplication modulo 7908.

Nathan Jacobson Few mathematicians have been as productive over such a long career or have had as much influence on the profession as has Professor Jacobson. Citation for the Steele Prize for Lifetime Achievement

NATHAN JACOBSON was born on September 8, 1910, in Warsaw, Poland. After arriving in the United States in 1917, Jacobson grew up in Alabama, Mississippi, and Georgia, where his father owned small clothing stores. He received a B.A. degree from the University of Alabama in 1930 and a Ph.D. from Princeton in 1934. After brief periods as a professor at Bryn Mawr, the University of Chicago, the University of North Carolina, and Johns Hopkins, Jacobson accepted a position at Yale, where he remained until his retirement in 1981. Jacobson’s principal contributions to algebra were in the areas of rings, Lie algebras, and Jordan algebras. In particular, he developed structure theories for these systems. He was the author of nine books and numerous articles, and he had 33 Ph.D. students.

Jacobson held visiting positions in France, India, Italy, Israel, China, Australia, and Switzerland. Among his many honors were the presidency of the American Mathematical Society, memberships in the National Academy of Sciences and the American Academy of Arts and Sciences, a Guggenheim Fellowship, and an honorary degree from the University of Chicago. Jacobson died on December 5, 1999, at the age of 89. To find more information about Jacobson, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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The secret of science is to ask the right questions, and it is the choice of problem more than anything else that marks the man of genius in the scientific world. SIR HENRY TIZARD IN C. P. SNOW, A postscript to Science and Government

Ideals Normal subgroups play a special role in group theory—they permit us to construct factor groups. In this chapter, we introduce the analogous concepts for rings—ideals and factor rings. Definition Ideal

A subring A of a ring R is called a (two-sided) ideal of R if for every r [ R and every a [ A both ra and ar are in A.

So, a subring A of a ring R is an ideal of R if A “absorbs” elements from R—that is, if rA 5 {ra| a [ A} # A and Ar 5 {ar| a [ A} # A for all r [ R. An ideal A of R is called a proper ideal of R if A is a proper subset of R. In practice, one identifies ideals with the following test, which is an immediate consequence of the definition of ideal and the subring test given in Theorem 12.3. Theorem 14.1 Ideal Test A nonempty subset A of a ring R is an ideal of R if 1. a 2 b [ A whenever a, b [ A. 2. ra and ar are in A whenever a [ A and r [ R.

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EXAMPLE 1 For any ring R, {0} and R are ideals of R. The ideal {0} is called the trivial ideal. EXAMPLE 2 For any positive integer n, the set nZ 5 {0, 6n, 62n, . . .} is an ideal of Z. EXAMPLE 3 Let R be a commutative ring with unity and let a [ R. The set a 5 {ra | r [ R} is an ideal of R called the principal ideal generated by a. (Notice that a is also the notation we used for the cyclic subgroup generated by a. However, the intended meaning will always be clear from the context.) The assumption that R is commutative is necessary in this example (see Exercise 29 in the Supplementary Exercises for Chapters 12–14). EXAMPLE 4 Let R[x] denote the set of all polynomials with real coefficients and let A denote the subset of all polynomials with constant term 0. Then A is an ideal of R[x] and A 5 x. EXAMPLE 5 Let R be a commutative ring with unity and let a1, a2, . . . , an belong to R. Then I 5 a1, a2, . . . , an 5 {r1a1 1 r2a2 1 ? ? ? 1 rnan | ri [ R} is an ideal of R called the ideal generated by a1, a2, . . . , an. The verification that I is an ideal is left as an easy exercise (Exercise 3). EXAMPLE 6 Let Z[x] denote the ring of all polynomials with integer coefficients and let I be the subset of Z[x] of all polynomials with even constant terms. Then I is an ideal of Z[x] and I 5 x, 2 (see Exercise 37). EXAMPLE 7 Let R be the ring of all real-valued functions of a real variable. The subset S of all differentiable functions is a subring of R but not an ideal of R.

Factor Rings Let R be a ring and let A be an ideal of R. Since R is a group under addition and A is a normal subgroup of R, we may form the factor group R/A 5 {r 1 A | r [ R}. The natural question at this point is: How may we form a ring of this group of cosets? The addition is already taken care of, and, by analogy with groups of cosets, we define the product of two cosets of s 1 A and t 1 A as st 1 A. The next theorem shows that this definition works as long as A is an ideal, and not just a subring, of R.

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Theorem 14.2 Existence of Factor Rings Let R be a ring and let A be a subring of R. The set of cosets {r 1 A | r [ R} is a ring under the operations (s 1 A) 1 (t 1 A) 5 s 1 t 1 A and (s 1 A)(t 1 A) 5 st 1 A if and only if A is an ideal of R.

PROOF We know that the set of cosets forms a group under addition. Once we know that multiplication is indeed a binary operation on the cosets, it is trivial to check that the multiplication is associative and that multiplication is distributive over addition. Hence, the proof boils down to showing that multiplication is well defined if and only if A is an ideal of R. To do this, let us suppose that A is an ideal and let s 1 A 5 s9 1 A and t 1 A 5 t9 1 A. Then we must show that st 1 A 5 s9t9 1 A. Well, by definition, s 5 s9 1 a and t 5 t9 1 b, where a and b belong to A. Then st 5 (s9 1 a)(t9 1 b) 5 s9t9 1 at9 1 s9b 1 ab, and so st 1 A 5 s9t9 1 at9 1 s9b 1 ab 1 A 5 s9t9 1 A, since A absorbs at9 1 s9b 1 ab. Thus, multiplication is well defined when A is an ideal. On the other hand, suppose that A is a subring of R that is not an ideal of R. Then there exist elements a [ A and r [ R such that ar o A or ra o A. For convenience, say ar o A. Consider the elements a 1 A 5 0 1 A and r 1 A. Clearly, (a 1 A)(r 1 A) 5 ar 1 A but (0 1 A) ? (r 1 A) 5 0 ? r 1 A 5 A. Since ar 1 A 2 A, the multiplication is not well defined and the set of cosets is not a ring. Let’s look at a few factor rings. EXAMPLE 8 Z/4Z 5 {0 1 4Z, 1 1 4Z, 2 1 4Z, 3 1 4Z}. To see how to add and multiply, consider 2 1 4Z and 3 1 4Z. (2 1 4Z) 1 (3 1 4Z) 5 5 1 4Z 5 1 1 4 1 4Z 5 1 1 4Z, (2 1 4Z)(3 1 4Z) 5 6 1 4Z 5 2 1 4 1 4Z 5 2 1 4Z. One can readily see that the two operations are essentially modulo 4 arithmetic.

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EXAMPLE 9 2Z/6Z 5 {0 1 6Z, 2 1 6Z, 4 1 6Z}. Here the operations are essentially modulo 6 arithmetic. For example, (4 1 6Z) 1 (4 1 6Z) 5 2 1 6Z and (4 1 6Z)(4 1 6Z) 5 4 1 6Z. Here is a noncommutative example of an ideal and factor ring. a1 a2 d ` a i P Z f and let I be the a3 a4 subset of R consisting of matrices with even entries. It is easy to show that I is indeed an ideal of R (Exercise 21). Consider the factor ring R/I. The interesting question about this ring is: What is its size? r r We claim R/I has 16 elements; in fact, R/I 5 ec 1 2 d1I 0 ri [ 50, 16f . r3 r4 An example illustrates the typical situation. Which of the 16 elements 7 8 7 8 1 0 is c d 1 I ? Well, observe that c d 1I5 c d1 5 23 5 23 1 1 6 8 1 0 c d 1I5 c d 1 I , since an ideal absorbs its own elements. 4 24 1 1 The general case is left to the reader (Exercise 23). EXAMPLE 10 Let R 5 e c

EXAMPLE 11 Consider the factor ring of the Gaussian integers R 5 Z[i]/2 2 i. What does this ring look like? Of course, the elements of R have the form a 1 bi 1 2 2 i, where a and b are integers, but the important question is: What do the distinct cosets look like? The fact that 2 2 i 1 2 2 i 5 0 1 2 2 i means that when dealing with coset representatives, we may treat 2 2 i as equivalent to 0, so that 2 5 i. For example, the coset 3 1 4i 1 2 2 i 5 3 1 8 1 2 2 i 5 11 1 2 2 i. Similarly, all the elements of R can be written in the form a 1 2 2 i, where a is an integer. But we can further reduce the set of distinct coset representatives by observing that when dealing with coset representatives, 2 5 i implies (by squaring both sides) that 4 5 21 or 5 5 0. Thus, the coset 3 1 4i 1 2 2 i 5 11 1 2 2 i 5 1 1 5 1 5 1 2 2 i 5 1 1 2 2 i. In this way, we can show that every element of R is equal to one of the following cosets: 0 1 2 2 i, 1 1 2 2 i, 2 1 2 2 i, 3 1 2 2 i, 4 1 2 2 i. Is any further reduction possible? To demonstrate that there is not, we will show that these five cosets are distinct. It suffices to show that 1 1 2 2 i has additive order 5. Since 5(1 1 2 2 i) 5 5 1 2 2 i 5 0 1 2 2 i, 1 1 2 2 i has order 1 or 5. If the order is actually 1, then 1 1 2 2 i 5 0 1 2 2 i, so 1 [ 2 2 i. Thus, 1 5 (2 2 i) (a 1 bi) 5 2a 1 b 1 (2a 1 2b)i for some integers a and b. But this equation implies that 1 5 2a 1 b and 0 5 2a 1 2b, and solving these

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simultaneously yields b 5 1/5, which is a contradiction. It should be clear that the ring R is essentially the same as the field Z5. EXAMPLE 12 Let R[x] denote the ring of polynomials with real coefficients and let x2 1 1 denote the principal ideal generated by x2 1 1; that is, x2 1 1 5 {f(x)(x2 1 1) | f(x) [ R[x]}. Then R[x]/x2 1 1 5 {g(x) 1 x2 1 1 | g(x) [ R[x]} 5 {ax 1 b 1 x2 1 1 | a, b [ R}. To see this last equality, note that if g(x) is any member of R[x], then we may write g(x) in the form q(x)(x2 1 1) 1 r(x), where q(x) is the quotient and r(x) is the remainder upon dividing g(x) by x2 1 1. In particular, r(x) 5 0 or the degree of r(x) is less than 2, so that r(x) 5 ax 1 b for some a and b in R. Thus, g(x) 1 x2 1 1 5 q(x)(x2 1 1) 1 r(x) 1 x2 1 1 5 r(x) 1 x2 1 1, since the ideal x2 1 1 absorbs the term q(x)(x2 1 1). How is multiplication done? Since x2 1 1 1 x2 1 1 5 0 1 x2 1 1, one should think of x2 1 1 as 0 or, equivalently, as x2 5 21. So, for example, (x 1 3 1 x2 1 1) ? (2x 1 5 1 x2 1 1) 5 2x2 1 11x 1 15 1 x2 1 1 5 11x 1 13 1 x2 1 1. In view of the fact that the elements of this ring have the form ax 1 b 1 x2 1 1, where x2 1 x2 1 1 5 21 1 x2 1 1, it is perhaps not surprising that this ring turns out to be algebraically the same ring as the ring of complex numbers. This observation was first made by Cauchy in 1847. Examples 11 and 12 illustrate one of the most important applications of factor rings—the construction of rings with highly desirable properties. In particular, we shall show how one may use factor rings to construct integral domains and fields.

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Prime Ideals and Maximal Ideals Definition Prime Ideal, Maximal Ideal

A prime ideal A of a commutative ring R is a proper ideal of R such that a, b [ R and ab [ A imply a [ A or b [ A. A maximal ideal of a commutative ring R is a proper ideal of R such that, whenever B is an ideal of R and A # B # R, then B 5 A or B 5 R.

So, the only ideal that properly contains a maximal ideal is the entire ring. The motivation for the definition of a prime ideal comes from the integers. EXAMPLE 13 Let n be an integer greater than 1. Then, in the ring of integers, the ideal nZ is prime if and only if n is prime (Exercise 9). ({0} is also a prime ideal of Z.) EXAMPLE 14 The lattice of ideals of Z36 (Figure 14.1) shows that only 2 and 3 are maximal ideals. EXAMPLE 15 The ideal x2 1 1 is maximal in R[x]. To see this, assume that A is an ideal of R[x] that properly contains x2 1 1. We will prove that A 5 R[x] by showing that A contains some nonzero real number c. [This is the constant polynomial h(x) 5 c for all x.] Then 1 5 (1/c)c [ A and therefore, by Exercise 15, A 5 R[x]. To this end, let f(x) [ A, but f(x) o x2 1 1. Then f(x) 5 q(x)(x2 1 1) 1 r(x), where r(x) 2 0 and the degree of r(x) is less than 2. It follows that r(x) 5 ax 1 b, where a and b are not both 0, and ax 1 b 5 r(x) 5 f(x) 2 q(x)(x2 1 1) [ A. Z 36 <2> <4>

<3> <6>

<9>

<18>

<12 >

<0>

Figure 14.1

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Thus, a2x2 2 b2 5 (ax 1 b)(ax 2 b) [ A

and

a2(x2 1 1) [ A.

So, 0 2 a2 1 b2 5 (a2x2 1 a2) 2 (a2x2 2 b2) [ A. EXAMPLE 16 The ideal x2 1 1 is not prime in Z2[x], since it contains (x 1 1)2 5 x2 1 2x 1 1 5 x2 1 1 but does not contain x 1 1. The next two theorems are useful for determining whether a particular ideal is prime or maximal. Theorem 14.3 R/A Is an Integral Domain If and Only If A Is Prime Let R be a commutative ring with unity and let A be an ideal of R. Then R/A is an integral domain if and only if A is prime.

PROOF Suppose that R/A is an integral domain and ab [ A. Then (a 1 A)(b 1 A) 5 ab 1 A 5 A, the zero element of the ring R/A. So, either a 1 A 5 A or b 1 A 5 A; that is, either a [ A or b [ A. Hence, A is prime. To prove the other half of the theorem, we first observe that R/A is a commutative ring with unity for any proper ideal A. Thus, our task is simply to show that when A is prime, R/A has no zero-divisors. So, suppose that A is prime and (a 1 A)(b 1 A) 5 0 1 A 5 A. Then ab [ A and, therefore, a [ A or b [ A. Thus, one of a 1 A or b 1 A is the zero coset in R/A. For maximal ideals, we can do even better. Theorem 14.4 R/A Is a Field If and Only If A Is Maximal Let R be a commutative ring with unity and let A be an ideal of R. Then R/A is a field if and only if A is maximal.

PROOF Suppose that R/A is a field and B is an ideal of R that properly contains A. Let b [ B but b o A. Then b 1 A is a nonzero element of R/A and, therefore, there exists an element c 1 A such that (b 1 A) ? (c 1 A) 5 1 1 A, the multiplicative identity of R/A. Since b [ B, we have bc [ B. Because 1 1 A 5 (b 1 A)(c 1 A) 5 bc 1 A,

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we have 1 2 bc [ A , B. So, 1 5 (1 2 bc) 1 bc [ B. By Exercise 15, B 5 R. This proves that A is maximal. Now suppose that A is maximal and let b [ R but b o A. It suffices to show that b 1 A has a multiplicative inverse. (All other properties for a field follow trivially.) Consider B 5 {br 1 a | r [ R, a [ A}. This is an ideal of R that properly contains A (Exercise 25). Since A is maximal, we must have B 5 R. Thus, 1 [ B, say, 1 5 bc 1 a9, where a9 [ A. Then 1 1 A 5 bc 1 a9 1 A 5 bc 1 A 5 (b 1 A)(c 1 A). When a commutative ring has a unity, it follows from Theorems 14.3 and 14.4 that a maximal ideal is a prime ideal. The next example shows that a prime ideal need not be maximal. EXAMPLE 17 The ideal x is a prime ideal in Z[x] but not a maximal ideal in Z[x]. To verify this, we begin with the observation that x 5 {f(x) [ Z[x] | f(0) 5 0} (see Exercise 29). Thus, if g(x)h(x) [ x, then g(0)h(0) 5 0. And since g(0) and h(0) are integers, we have g(0) 5 0 or h(0) 5 0. To see that x is not maximal, we simply note that x , x, 2 , Z[x] (see Exercise 37).

Exercises Text not available due to copyright restrictions

1. Verify that the set defined in Example 3 is an ideal. 2. Verify that the set A in Example 4 is an ideal and that A 5 x. 3. Verify that the set I in Example 5 is an ideal and that if J is any ideal of R that contains a1, a2, . . . , an, then I # J. (Hence, a1, a2, . . . , an is the smallest ideal of R that contains a1, a2, . . . , an .) 4. Find a subring of Z % Z that is not an ideal of Z % Z. 5. Let S 5 {a 1 bi | a, b [ Z, b is even}. Show that S is a subring of Z[i], but not an ideal of Z[i]. 6. Find all maximal ideals in a. Z8, b. Z10, c. Z12, d. Zn. 7. Let a belong to a commutative ring R. Show that aR 5 {ar | r [ R} is an ideal of R. If R is the ring of even integers, list the elements of 4R. 8. Prove that the intersection of any set of ideals of a ring is an ideal.

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9. If n is an integer greater than 1, show that n 5 nZ is a prime ideal of Z if and only if n is prime. (This exercise is referred to in this chapter.) 10. If A and B are ideals of a ring, show that the sum of A and B, A 1 B 5 {a 1 b | a [ A, b [ B}, is an ideal. 11. In the ring of integers, find a positive integer a such that a. a 5 2 1 3, b. a 5 6 1 8, c. a 5 m 1 n. 12. If A and B are ideals of a ring, show that the product of A and B, AB 5 {a1b1 1 a2b2 1 ? ? ? 1 anbn | ai [ A, bi [ B, n a positive integer}, is an ideal. 13. Find a positive integer a such that a. a 5 34, b. a 5 68, c. a 5 mn. 14. Let A and B be ideals of a ring. Prove that AB # A > B. 15. If A is an ideal of a ring R and 1 belongs to A, prove that A 5 R. (This exercise is referred to in this chapter.) 16. If A and B are ideals of a commutative ring R with unity and A 1 B 5 R, show that A > B 5 AB. 17. If an ideal I of a ring R contains a unit, show that I 5 R. 18. Suppose that in the ring Z the ideal 35 is a proper ideal of J and J is a proper ideal of I. What are the possibilities for J? What are the possibilities for I? 19. Give an example of a ring that has exactly two maximal ideals. 20. Suppose that R is a commutative ring and |R| 5 30. If I is an ideal of R and |I| 5 10, prove that I is a maximal ideal. 21. Let R and I be as described in Example 10. Prove that I is an ideal of R. 22. Let I 5 2. Prove that I[x] is not a maximal ideal of Z[x] even though I is a maximal ideal of Z. 23. Verify the claim made in Example 10 about the size of R/I. 24. Give an example of a commutative ring that has a maximal ideal that is not a prime ideal. 25. Show that the set B in the latter half of the proof of Theorem 14.4 is an ideal of R. (This exercise is referred to in this chapter.) 26. If R is a commutative ring with unity and A is a proper ideal of R, show that R/A is a commutative ring with unity. 27. Prove that the only ideals of a field F are {0} and F itself.

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271

28. Show that R[x]/x2 1 1 is a field. 29. In Z[x], the ring of polynomials with integer coefficients, let I 5 {f (x) [ Z[x] | f(0) 5 0}. Prove that I 5 x. (This exercise is referred to in this chapter and in Chapter 15.) 30. Show that A 5 {(3x, y) | x, y [ Z} is a maximal ideal of Z % Z. Generalize. What happens if 3x is replaced by 4x? Generalize. 31. Let R be the ring of continuous functions from R to R. Show that A 5 {f [ R | f(0) 5 0} is a maximal ideal of R. 32. Let R 5 Z8 % Z30. Find all maximal ideals of R, and for each maximal ideal I, identify the size of the field R/I. 33. How many elements are in Z[i]/3 1 i? Give reasons for your answer. 34. In Z[x], the ring of polynomials with integer coefficients, let I 5 { f(x) [ Z[x] | f (0) 5 0}. Prove that I is not a maximal ideal. 35. In Z % Z, let I 5 {(a, 0) | a [ Z}. Show that I is a prime ideal but not a maximal ideal. 36. Let R be a ring and let I be an ideal of R. Prove that the factor ring R/I is commutative if and only if rs 2 sr [ I for all r and s in R. 37. In Z[x], let I 5 { f(x) [ Z[x] | f (0) is an even integer}. Prove that I 5 x, 2. Is I a prime ideal of Z[x]? Is I a maximal ideal? How many elements does Z[x]/I have? (This exercise is referred to in this chapter.) 38. Prove that I 5 2 1 2i is not a prime ideal of Z[i]. How many elements are in Z[i]/I? What is the characteristic of Z[i]/I? 39. In Z5[x], let I 5 x2 1 x 1 2. Find the multiplicative inverse of 2x 1 3 1 I in Z5[x]/I. 40. Let R be a ring and let p be a fixed prime. Show that Ip 5 {r [ R | additive order of r is a power of p} is an ideal of R. 41. An integral domain D is called a principal ideal domain if every ideal of D has the form a 5 {ad | d [ D} for some a in D. Show that Z is a principal ideal domain. (This exercise is referred to in Chapter 18.) 42. Let R 5 e c

a 0

b r d 0 a, b, d [ Z6 and S 5 e c d 0

s d 0 r, s, t [ Z, s t

is even f . If S is an ideal of R, what can you say about r and t?

43. If R and S are principal ideal domains, prove that R % S is a principal ideal ring.

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44. Let a and b belong to a commutative ring R. Prove that {x [ R | ax [ bR} is an ideal. 45. Let R be a commutative ring and let A be any subset of R. Show that the annihilator of A, Ann(A) 5 {r [ R | ra 5 0 for all a in A}, is an ideal. 46. Let R be a commutative ring and let A be any ideal of R. Show that the nil radical of A, N(A) 5 {r [ R | r n [ A for some positive integer n (n depends on r)}, is an ideal of R. [N(0) is called the nil radical of R.] 47. Let R 5 Z27. Find a. N(0), b. N(3), c. N(9). 48. Let R 5 Z36. Find a. N(0), b. N(4), c. N(6). 49. Let R be a commutative ring. Show that R/N(0) has no nonzero nilpotent elements. 50. Let A be an ideal of a commutative ring. Prove that N(N(A)) 5 N(A). 51. Let Z2[x] be the ring of all polynomials with coefficients in Z2 (that is, coefficients are 0 or 1, and addition and multiplication of coefficients are done modulo 2). Show that Z2[x]/x2 1 x 1 1 is a field. 52. List the elements of the field given in Exercise 51, and make an addition and multiplication table for the field. 53. Show that Z 3[x]/x2 1 x 1 1 is not a field. 54. Let R be a commutative ring without unity, and let a [ R. Describe the smallest ideal I of R that contains a (that is, if J is any ideal that contains a, then I # J). 55. Let R be the ring of continuous functions from R to R. Let A 5 { f [ R | f (0) is an even integer}. Show that A is a subring of R, but not an ideal of R. 56. Show that Z[i]/1 2 i is a field. How many elements does this field have? 57. If R is a principal ideal domain and I is an ideal of R, prove that every ideal of R/I is principal (see Exercise 41). 58. How many elements are in Z5[i]/1 1 i? 59. Let R be a commutative ring with unity that has the property that a2 5 a for all a in R. Let I be a prime ideal in R. Show that |R/I| 5 2. 60. Let R be a commutative ring with unity, and let I be a proper ideal with the property that every element of R that is not in I is a unit of R. Prove that I is the unique maximal ideal of R.

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61. Let I0 5 { f(x) [ Z[x] | f(0) 5 0}. For any positive integer n, show that there exists a sequence of strictly increasing ideals such that I0 , I1 , I2 , ? ? ? , In , Z[x]. 62. Let R 5 {(a1, a2, a3, . . .)}, where each ai [ Z. Let I 5 {(a1, a2, a3, . . . )}, where only a finite number of terms are nonzero. Prove that I is not a principal ideal of R. 63. Let R be a commutative ring with unity and let a, b [ R. Show that a, b, the smallest ideal of R containing a and b, is I 5 {ra 1 sb| r, s [ R}. That is, show that I contains a and b and that any ideal that contains a and b also contains I.

Computer Exercises What is the common denominator of intellectual accomplishment? In math, science, economics, history, or any other subject, the answer is the same: great thinkers notice patterns. DAVID NIVEN, PSYCHOLOGIST

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software determines the number of elements in the ring Z[i]/ a 1 bi (where i2 5 21). Run the program for several cases and formulate a conjecture based on your data. 2. This software determines the characteristic of the ring Z[i]/a 1 bi (where i2 5 21). Let d 5 gcd(a, b). Run the program for several cases with d 5 1 and formulate a conjecture based on your data. Run the program for several cases with d . 1 and formulate a conjecture in terms of a, b, and d based on your data. Does the formula you found for d . 1 also work in the case that d 5 1? 3. This software determines when the ring Z[i]/a 1 bi (where i2 5 21) is isomorphic to the ring Za21b2. Run the program for several cases and formulate a conjecture based on your data.

Richard Dedekind Richard Dedekind was not only a mathematician, but one of the wholly great in the history of mathematics, now and in the past, the last hero of a great epoch, the last pupil of Gauss, for four decades himself a classic, from whose works not only we, but our teachers and the teachers of our teachers, have drawn. EDMUND LANDAU, Commemorative Address to the Royal Society of Göttingen

This stamp was issued by East Germany in 1981 to commemorate the 150th anniversary of Dedekind’s birth. Notice that it features the representation of an ideal as the product of powers of prime ideals.

RICHARD DEDEKIND was born on October 6, 1831, in Brunswick, Germany, the birthplace of Gauss. Dedekind was the youngest of four children of a law professor. His early interests were in chemistry and physics, but he obtained a doctor’s degree in mathematics at the age of 21 under Gauss at the University of Göttingen. Dedekind continued his studies at Göttingen for a few years, and in 1854 he began to lecture there.

274

Dedekind spent the years 1858–1862 as a professor in Zürich. Then he accepted a position at an institute in Brunswick where he had once been a student. Although this school was less than university level, Dedekind remained there for the next 50 years. He died in Brunswick in 1916. During his career, Dedekind made numerous fundamental contributions to mathematics. His treatment of irrational numbers, “Dedekind cuts,” put analysis on a firm, logical foundation. His work on unique factorization led to the modern theory of algebraic numbers. He was a pioneer in the theory of rings and fields. The notion of ideals as well as the term itself are attributed to Dedekind. Mathematics historian Morris Kline has called him “the effective founder of abstract algebra.” To find more information about Dedekind, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Emmy Noether In the judgment of the most competent living mathematicians, Fräulein Noether was the most significant creative mathematical genius thus far produced since the higher education of women began. In the realm of algebra, in which the most gifted mathematicians have been busy for centuries, she discovered methods which have proved of enormous importance in the development of the present-day younger generation of mathematicians. ALBERT EINSTEIN,

The New York Times

EMMY NOETHER was born on March 23, 1882, in Germany. When she entered the University of Erlangen, she was one of only two women among the 1000 students. Noether completed her doctorate in 1907. In 1916, Noether went to Göttingen and, under the influence of David Hilbert and Felix Klein, became interested in general relativity. While there, she made a major contribution to physics with her theorem that whenever there is a symmetry in nature, there is also a conservation law, and vice versa. Hilbert tried unsuccessfully to obtain a faculty appointment at Göttingen for Noether, saying, “I do not see that the sex of the candidate is an argument against her admission as Privatdozent. After all, we are a university and not a bathing establishment.”

It was not until she was 38 that Noether’s true genius revealed itself. Over the next 13 years, she used an axiomatic method to develop a general theory of ideals and noncommutative algebras. With this abstract theory, Noether was able to weld together many important concepts. Her approach was even more important than the individual results. Hermann Weyl said of Noether, “She originated above all a new and epoch-making style of thinking in algebra.” With the rise of Hitler in 1933, Noether, a Jew, fled to the United States and took a position at Bryn Mawr College. She died suddenly on April 14, 1935, following an operation. To find more information about Noether, visit: http://www-groups.dcs .st-and.ac.uk/~history/

275

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Supplementary Exercises for Chapters 12–14 If at first you do succeed—try to hide your astonishment. HARRY F. BANKS

True/false questions for Chapters 12–14 are available on the Web at: http://www.d.umn.edu/~jgallian/TF 1. Find all idempotent elements in Z10, Z20, and Z30. (Recall that a is idempotent if a2 5 a.) 2. If m and n are relatively prime integers greater than 1, prove that Zmn has at least two idempotents besides 0 and 1. 3. Suppose that R is a ring in which a2 5 0 implies a 5 0. Show that R has no nonzero nilpotent elements. (Recall that b is nilpotent if bn 5 0 for some positive integer n.) 4. Let R be a commutative ring with more than one element. Prove that if for every nonzero element a of R we have aR 5 R, then R is a field. 5. Let A, B, and C be ideals of a ring R. If AB # C and C is a prime ideal of R, show that A # C or B # C. (Compare this with Euclid’s Lemma in Chapter 0.) 6. Show, by example, that the intersection of two prime ideals need not be a prime ideal. 7. Let R denote the ring of real numbers. Determine all ideals of R % R. What happens if R is replaced by any field F? 8. Determine all factor rings of Z. 9. Suppose that n is a square-free positive integer (that is, n is not divisible by the square of any prime). Prove that Zn has no nonzero nilpotent elements. 10. Let R be a commutative ring with unity. Suppose that a is a unit and b is nilpotent. Show that a 1 b is a unit. (Hint: See Exercise 29 in Chapter 12.) 11. Let A, B, and C be subrings of a ring R. If A # B < C, show that A # B or A # C. 12. For any element a in a ring R, define a to be the smallest ideal of R that contains a. If R is a commutative ring with unity, show that a 5 aR 5 {ar | r [ R}. Show, by example, that if R is commutative but does not have a unity, then a and aR may be different. 13. Let R be a ring with unity. Show that a 5 {s1at1 1 s2at2 1 ? ? ? 1 snatn | si, ti [ R and n is a positive integer}. 14. Show that Zn[x] has characteristic n.

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277

15. Let A and B be ideals of a ring R. If A > B 5 {0}, show that ab 5 0 when a [ A and b [ B. 16. Show that the direct sum of two integral domains is not an integral domain. 17. Consider the ring R 5 {0, 2, 4, 6, 8, 10} under addition and multiplication modulo 12. What is the characteristic of R? 18. What is the characteristic of Zm % Zn? Generalize. 19. Let R be a commutative ring with unity. Suppose that the only ideals of R are {0} and R. Show that R is a field. 20. Suppose that I is an ideal of J and that J is an ideal of R. Prove that if I has a unity, then I is an ideal of R. (Be careful not to assume that the unity of I is the unity of R. It need not be—see Exercise 2 in Chapter 12.) 21. Recall that an idempotent element b in a ring is one with the property that b2 5 b. Find a nontrivial idempotent (that is, not 0 and not 1) in Q[x]/x4 1 x2. 22. In a principal ideal domain, show that every nontrivial prime ideal is a maximal ideal. 23. Find an example of a commutative ring R with unity such that a, b [ R, a 2 b, and an 5 bn and am 5 bm, where n and m are positive integers that are relatively prime. (Compare with Exercise 35, part b, in Chapter 13.) 3 24. Let Q( " 2) denote the smallest subfield of R that contains Q and 3 3 3 "2. [That is, Q( "2) is the subfield with the property that Q( "2) 3 3 contains Q and "2 and if F is any subfield containing Q and " 2, 3 3 then F contains Q( "2).] Describe the elements of Q( "2). 25. Let R be an integral domain with nonzero characteristic. If A is a proper ideal of R, show that R/A has the same characteristic as R. 26. Let F be a field of order pn. Determine the group isomorphism class of F under the operation addition. 27. If R is a finite commutative ring with unity, prove that every prime ideal of R is a maximal ideal of R. 28. Let R be a noncommutative ring and let C(R) be the center of R (see Exercise 19 in Chapter 12). Prove that the additive group of R/C(R) is not cyclic. 29. Let R5 ec

a b d ` a, b, c, d [ Z 2 f c d

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with ordinary matrix addition and multiplication modulo 2. Show that A5 ec

30. 31. 32.

33. 34. 35.

36. 37. 38. 39. 40.

41. 42. 43. 44.

1 0 dr ` r [ Rf 0 0

is not an ideal of R. (Hence, in Exercise 7 in Chapter 14, the commutativity assumption is necessary.) If R is an integral domain and A is a proper ideal of R, must R/A be an integral domain? Let A 5 {a 1 bi | a, b [ Z, a mod 2 5 b mod 2}. How many elements does Z[i]/A have? Show that A is a maximal ideal of Z[i]. Suppose that R is a commutative ring with unity such that for each a in R there is a positive integer n greater than 1 (n depends on a) such that an 5 a. Prove that every prime ideal of R is a maximal ideal of R. State a “finite subfield test”; that is, state conditions that guarantee that a finite subset of a field is a subfield. Let F be a finite field with more than two elements. Prove that the sum of all of the elements of F is 0. Show that if there are nonzero elements a and b in Zn such that a2 1 b2 5 0, then the ring Zn[i] 5 {x 1 yi | x, y [ Zn} has zero-divisors. Use this fact to find a zero-divisor in Z13[i]. Suppose that R is a ring with no zero-divisors and that R contains a nonzero element b such that b2 5 b. Show that b is a unity for R. Find the characteristic of Z[i]/2 1 i. Show that the characteristic of Z[i]/a 1 bi divides a2 1 b2. Show that 4x2 1 6x 1 3 is a unit in Z8[x]. For any commutative ring R, R[x, y] is the ring of polynomials in x and y with coefficients in R (that is, R[x, y] consists of all finite sums of terms of the form axiyj, where a [ R and i and j are nonnegative integers). (For example, x4 2 3x2y 2 y3 [ Z[x, y].) Prove that x, y is a prime ideal in Z[x, y] but not a maximal ideal in Z[x, y]. Prove that x, y is a maximal ideal in Z5[x, y]. Prove that 2, x, y is a maximal ideal in Z[x, y]. Let R and S be rings. Prove that (a, b) is nilpotent in R % S if and only if both a and b are nilpotent. Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in R % S if and only if a or b is a zero-divisor or exactly one of a or b is 0.

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45. Determine all idempotents in Zp k, where p is a prime. 46. Let R be a commutative ring with unity 1. Show that a is an idempotent if and only if there exists an element b in R such that ab 5 0 and a 1 b 5 1. 47. Let Zn["2] 5 {a 1 b"2| a, b [ Zn}. Define addition and multiplication as in Z["2], except that modulo n arithmetic is used to combine the coefficients. Show that Z3["2] is a field but Z7["2] is not. 48. Let p be a prime. Prove that every zero-divisor in Z pn is a nilpotent element. 49. If x is a nilpotent element in a commutative ring R, prove that rx is nilpotent for all r in R.

15 Ring Homomorphisms

If there is one central idea which is common to all aspects of modern algebra it is the notion of homomorphism. I. N. HERSTEIN, Topics in Algebra

Definition and Examples In our work with groups, we saw that one way to discover information about a group is to examine its interaction with other groups by way of homomorphisms. It should not be surprising to learn that this concept extends to rings with equally profitable results. Just as a group homomorphism preserves the group operation, a ring homomorphism preserves the ring operations. Definitions Ring Homomorphism, Ring Isomorphism

A ring homomorphism f from a ring R to a ring S is a mapping from R to S that preserves the two ring operations; that is, for all a, b in R,

f(a 1 b) 5 f(a) 1 f(b)

and

f(ab) 5 f(a)f(b).

A ring homomorphism that is both one-to-one and onto is called a ring isomorphism.

As is the case for groups, in the preceding definition the operations on the left of the equal signs are those of R, whereas the operations on the right of the equal signs are those of S. Again as with group theory, the roles of isomorphisms and homomorphisms are entirely distinct. An isomorphism is used to show that two rings are algebraically identical; a homomorphism is used to simplify a ring while retaining certain of its features. A schematic representation of a ring homomorphism is given in Figure 15.1. The dashed arrows indicate the results of performing the ring operations. The following examples illustrate ring homomorphisms. The reader should supply the missing details. 280

15 | Ring Homomorphisms

R

φ

b

φ

a

a1b

S

φ (b) φ (a)

φ

a.b

φ

281

φ (a) . φ (b) φ (a) 1 φ (b)

Figure 15.1

EXAMPLE 1 For any positive integer n, the mapping k S k mod n is a ring homomorphism from Z onto Zn (see Exercise 11 in Chapter 0). This mapping is called the natural homomorphism from Z to Zn. EXAMPLE 2 The mapping a 1 bi S a 2 bi is a ring isomorphism from the complex numbers onto the complex numbers (see Exercise 25 in Chapter 6). EXAMPLE 3 Let R[x] denote the ring of all polynomials with real coefficients. The mapping f(x) S f(1) is a ring homomorphism from R[x] onto R. EXAMPLE 4 The correspondence f: x S 5x from Z4 to Z10 is a ring homomorphism. Although showing that f(x 1 y) 5 f(x) 1 f(y) appears to be accomplished by the simple statement that 5(x 1 y) 5 5x 1 5y, we must bear in mind that the addition on the left is done modulo 4, whereas the addition on the right and the multiplication on both sides are done modulo 10. An analogous difficulty arises in showing that f preserves multiplication. So, to verify that f preserves both operations, we write x 1 y 5 4q1 1 r1 and xy 5 4q2 1 r2, where 0 # r1 , 4 and 0 # r2 , 4. Then f(x 1 y) 5 f(r1) 5 5r1 5 5(x 1 y 2 4q1) 5 5x 1 5y 2 20q1 5 5x 1 5y 5 f(x) 1 f(y) in Z10. Similarly, using the fact that 5 ? 5 5 5 in Z10, we have f(xy) 5 f(r2) 5 5r2 5 5(xy 2 4q2) 5 5xy 2 20q2 5 (5 ? 5)xy 5 5x5y 5 f(x)f(y) in Z10. EXAMPLE 5 We determine all ring homomorphisms from Z12 to Z30. By Example 10 in Chapter 10, the only group homomorphisms from Z12 to Z30 are x S ax, where a 5 0, 15, 10, 20, 5, or 25. But, since 1 ? 1 5 1 in Z12, we must have a ? a 5 a in Z30. This requirement rules out 20 and 5 as possibilities for a. Finally, simple calculations show that each of the remaining four choices does yield a ring homomorphism.

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EXAMPLE 6 Let R be a commutative ring of characteristic 2. Then the mapping a S a2 is a ring homomorphism from R to R. EXAMPLE 7 Although 2Z, the group of even integers under addition, is group-isomorphic to the group Z under addition, the ring 2Z is not ring-isomorphic to the ring Z. (Quick! What does Z have that 2Z doesn’t?) Our next two examples are applications to number theory of the natural homomorphism given in Example 1. EXAMPLE 8 (Test for Divisibility by 9) An integer n with decimal representation akak21 ? ? ? a0 is divisible by 9 if and only if ak 1 ak21 1 ? ? ? 1 a0 is divisible by 9. To verify this, observe that n 5 ak10k 1 ak2110k21 1 ? ? ? 1 a0. Then, letting a denote the natural homomorphism from Z to Z9 [in particular, a(10) 5 1], we note that n is divisible by 9 if and only if 0 5 a(n) 5 a(ak)(a(10))k 1 a(ak21)(a(10))k21 1 ? ? ? 1 a(a0) 5 a(ak) 1 a(ak21) 1 ? ? ? 1 a(a0) 5 a(ak 1 ak21 1 ? ? ? 1 a0). But a(ak 1 ak21 1 ? ? ? 1 a0) 5 0 is equivalent to ak 1 ak21 1 ? ? ? 1 a0 being divisible by 9. EXAMPLE 9 (Theorem of Gersonides) Among the most important unsolved problems in number theory is the so-called “abc conjecture.” This conjecture is a natural generalization of a theorem first proved in the fourteenth century by the Rabbi Gersonides. Gersonides proved that the only pairs of positive integers that are powers of 2 and powers of 3 which differ by 1 are 1, 2; 2, 3; 3, 4; and 8, 9. That is, these four pairs are the only solutions to the equations 2m 5 3n 6 1. To verify that this is so for 2m 5 3n 1 1, observe that for all n we have 3n mod 8 5 3 or 1. Thus, 3n 1 1 mod 8 5 4 or 2. On the other hand, for m . 3, we have 2m mod 8 5 0. To handle the case where 2m 5 3n 2 1, we first note that for all n, 3n mod 16 5 3, 9, 11, or 1, depending on the value of n mod 4. Thus, (3n 2 1) mod 16 5 2, 8, 10, or 0. Since 2m mod 16 5 0 for m $ 4, we have ruled out the cases where n mod 4 5 1, 2, or 3. Because 34k mod 5 5 (34)k mod 5 5 1k mod 5 5 1, we know that (34k 2 1) mod 5 5 0. But the only values for 2m mod 5 are 2, 4, 3, and 1. This contradiction completes the proof.

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Properties of Ring Homomorphisms Theorem 15.1 Properties of Ring Homomorphisms Let f be a ring homomorphism from a ring R to a ring S. Let A be a subring of R and let B be an ideal of S. 1. For any r [ R and any positive integer n, f(nr) 5 nf(r) and f(r n) 5 (f(r))n. 2. f(A) 5 {f(a) | a [ A} is a subring of S. 3. If A is an ideal and f is onto S, then f(A) is an ideal. 4. f21(B) 5 {r [ R | f(r) [ B} is an ideal of R. 5. If R is commutative, then f(R) is commutative. 6. If R has a unity 1, S 2 {0}, and f is onto, then f(1) is the unity of S. 7. f is an isomorphism if and only if f is onto and Ker f 5 {r [ R | f(r) 5 0} 5 {0}. 8. If f is an isomorphism from R onto S, then f21 is an isomorphism from S onto R.

PROOF The proofs of these properties are similar to those given in Theorems 10.1 and 10.2 and are left as exercises (Exercise 1). The student should learn the various properties of Theorem 15.1 in words in addition to the symbols. Property 2 says that the homomorphic image of a subring is a subring. Property 4 says that the pullback of an ideal is an ideal, and so on. The next three theorems parallel results we had for groups. The proofs are nearly identical to their group theory counterparts and are left as exercises (Exercises 2, 3, and 4). Theorem 15.2 Kernels Are Ideals Let f be a ring homomorphism from a ring R to a ring S. Then Ker f 5 {r [ R | f(r) 5 0} is an ideal of R.

Theorem 15.3 First Isomorphism Theorem for Rings Let f be a ring homomorphism from R to S. Then the mapping from R/Ker f to f(R), given by r 1 Ker f S f(r), is an isomorphism. In symbols, R/Ker f < f(R).

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Theorem 15.4 Ideals Are Kernels Every ideal of a ring R is the kernel of a ring homomorphism of R. In particular, an ideal A is the kernel of the mapping r S r 1 A from R to R/A.

The homomorphism from R to R/A given in Theorem 15.4 is called the natural homomorphism from R to R/A. Theorem 15.3 is often referred to as the Fundamental Theorem of Ring Homomorphisms. In Example 17 in Chapter 14 we gave a direct proof that x is a prime ideal of Z[x] but not a maximal ideal. In the following example we illustrate a better way to do this kind of problem. EXAMPLE 10 Since the mapping f from Z[x] onto Z given by f( f(x)) 5 f(0) is a ring homomorphism with Ker f 5 x (see Exercise 29 in Chapter 14), we have, by Theorem 15.3, Z[x]/x < Z. And because Z is an integral domain but not a field, we know by Theorems 14.3 and 14.4 that the ideal x is prime but not maximal in Z[x]. Theorem 15.5 Homomorphism from Z to a Ring with Unity Let R be a ring with unity 1. The mapping f: Z S R given by n S n ? 1 is a ring homomorphism.

PROOF Since the multiplicative group property am+n 5 aman translates to (m 1 n)a 5 ma 1 na when the operation is addition, we have f(m 1 n) 5 (m 1 n) ? 1 5 m ? 1 1 n ? 1. So, f preserves addition. That f also preserves multiplication follows from Exercise 15 in Chapter 12, which says that (m ? a)(n ? b) 5 (mn) ? (ab) for all integers m and n. Thus, f(mn) 5 (mn) ? 1 5 (mn) ? ((1)(1)) 5 (m ? 1)(n ? 1) 5 f(m)f(n). So, f preserves multiplication as well. Corollary 1 A Ring with Unity Contains Zn or Z If R is a ring with unity and the characteristic of R is n . 0, then R contains a subring isomorphic to Zn. If the characteristic of R is 0, then R contains a subring isomorphic to Z.

PROOF Let 1 be the unity of R and let S 5 {k ? 1 | k [ Z}. Theorem 15.5 shows that the mapping f from Z to S given by f(k) 5 k ? 1 is a homomorphism, and by the First Isomorphism Theorem for rings, we have Z/Ker f < S. But, clearly, Ker f 5 n, where n is the additive order of 1

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and, by Theorem 13.3, n is also the characteristic of R. So, when R has characteristic n, S < Z/n < Zn. When R has characteristic 0, S < Z/0 < Z. Corollary 2 Zm Is a Homomorphic Image of Z For any positive integer m, the mapping of f: Z S Zm given by x S x mod m is a ring homomorphism.

PROOF This follows directly from the statement of Theorem 15.5, since in the ring Zm, the integer x mod m is x ? 1. (For example, in Z3, if x 5 5, we have 5 ? 1 5 1 1 1 1 1 1 1 1 1 5 2.) Corollary 3 A Field Contains Zp or Q (Steinitz, 1910) If F is a field of characteristic p, then F contains a subfield isomorphic to Zp. If F is a field of characteristic 0, then F contains a subfield isomorphic to the rational numbers.

PROOF By Corollary 1, F contains a subring isomorphic to Zp if F has characteristic p, and F has a subring S isomorphic to Z if F has characteristic 0. In the latter case, let T 5 {ab21 | a, b [ S, b 2 0}. Then T is isomorphic to the rationals (Exercise 63). Since the intersection of all subfields of a field is itself a subfield (Exercise 11), every field has a smallest subfield (that is, a subfield that is contained in every subfield). This subfield is called the prime subfield of the field. It follows from Corollary 3 that the prime subfield of a field of characteristic p is isomorphic to Zp, whereas the prime subfield of a field of characteristic 0 is isomorphic to Q. (See Exercise 67.)

The Field of Quotients Although the integral domain Z is not a field, it is at least contained in a field—the field of rational numbers. And notice that the field of rational numbers is nothing more than quotients of integers. Can we mimic the construction of the rationals from the integers for other integral domains? Yes. The field constructed in Theorem 15.6 is called the field of quotients of D. Throughout the proof of Theorem 15.6, you should keep

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in mind that we are using the construction of the rationals from the integers as a model for our construction of the field of quotients of D. Theorem 15.6 Field of Quotients Let D be an integral domain. Then there exists a field F (called the field of quotients of D) that contains a subring isomorphic to D.

PROOF Let S 5 {(a, b) | a, b [ D, b 2 0}. We define an equivalence relation on S by (a, b) ; (c, d ) if ad 5 bc (compare with Example 14 in Chapter 0). Now, let F be the set of equivalence classes of S under the relation ; and denote the equivalence class that contains (x, y) by x/y. We define addition and multiplication on F by a/b 1 c/d 5 (ad 1 bc)/(bd )

and

a/b ? c/d 5 (ac)/(bd).

(Notice that here we need the fact that D is an integral domain to ensure that multiplication is closed; that is, bd 2 0 whenever b 2 0 and d 2 0.) Since there are many representations of any particular element of F (just as in the rationals, we have 1/2 5 3/6 5 4/8), we must show that these two operations are well defined. To do this, suppose that a/b 5 a9/b9 and c/d 5 c9/d9, so that ab9 5 a9b and cd9 5 c9d. It then follows that (ad 1 bc)b9d9 5 adb9d9 1 bcb9d9 5 (ab9)dd9 1 (cd9)bb9 5 (a9b)dd9 1 (c9d)bb9 5 a9d9bd 1 b9c9bd 5 (a9d9 1 b9c9)bd. Thus, by definition, we have (ad 1 bc)/(bd) 5 (a9d9 1 b9c9)/(b9d9), and, therefore, addition is well defined. We leave the verification that multiplication is well defined as an exercise (Exercise 55). That F is a field is straightforward. Let 1 denote the unity of D. Then 0/1 is the additive identity of F. The additive inverse of a/b is 2a/b; the multiplicative inverse of a nonzero element a/b is b/a. The remaining field properties can be checked easily. Finally, the mapping f: D S F given by x S x/1 is a ring isomorphism from D to f(D) (see Exercise 7). EXAMPLE 11 Let D 5 Z[x]. Then the field of quotients of D is {f(x)/g(x) | f(x), g(x) [ D, where g(x) is not the zero polynomial}. When F is a field, the field of quotients of F[x] is traditionally denoted by F(x).

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EXAMPLE 12 Let p be a prime. Then Zp(x) 5 {f(x)/g(x) | f(x), g(x) [ Zp[x], g(x) 2 0} is an infinite field of characteristic p.

Exercises We can work it out. TITLE OF SONG BY JOHN LENNON AND PAUL MCCARTNEY, December

1. 2. 3. 4. 5. 6. 7. 8. 9.

10. 11. 12. 13.

1965

Prove Theorem 15.1. Prove Theorem 15.2. Prove Theorem 15.3. Prove Theorem 15.4. Show that the correspondence x S 5x from Z5 to Z10 does not preserve addition. Show that the correspondence x S 3x from Z4 to Z12 does not preserve multiplication. Show that the mapping f: D S F in the proof of Theorem 15.6 is a ring homomorphism. Prove that every ring homomorphism f from Zn to itself has the form f(x) 5 ax, where a2 5 a. Suppose that f is a ring homomorphism from Zm to Zn. Prove that if f(1) 5 a, then a2 5 a. Give an example to show that the converse is false. a. Is the ring 2Z isomorphic to the ring 3Z? b. Is the ring 2Z isomorphic to the ring 4Z? Prove that the intersection of any collection of subfields of a field F is a subfield of F. (This exercise is referred to in this chapter.) Let Z3[i] 5 {a 1 bi | a, b [ Z3} (see Example 9 in Chapter 13). Show that the field Z3[i] is ring-isomorphic to the field Z3[x]/x2 1 1. Let a b d `a, b [ R f. 2b a Show that f: C S S given by a b f(a 1 bi) 5 c d 2b a is a ring isomorphism. S5ec

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14. Let Z["2] 5 {a 1 b"2 | a, b [ Z}. Let H5 ec

a 2b d ` a, b [ Z f . b a

Show that Z["2] and H are isomorphic as rings.

15. Consider the mapping from M2(Z ) into Z given by c

a c

b d S a. d

Prove or disprove that this is a ring homomorphism. 16. Let R 5 e c

18. 19. 20. 21. 22. 23. 24. 25. 26.

27. 28. 29. 30. 31.

b d ` a, b, c [ Z f . Prove or disprove that the mapc

a b d S a is a ring homomorphism. 0 c Is the mapping from Z5 to Z30 given by x S 6x a ring homomorphism? Note that the image of the unity is the unity of the image but not the unity of Z30. Is the mapping from Z10 to Z10 given by x S 2x a ring homomorphism? Describe the kernel of the homomorphism given in Example 3. Recall that a ring element a is called an idempotent if a2 5 a. Prove that a ring homomorphism carries an idempotent to an idempotent. Determine all ring homomorphisms from Z6 to Z6. Determine all ring homomorphisms from Z20 to Z30. Determine all ring isomorphisms from Zn to itself. Determine all ring homomorphisms from Z to Z. Suppose f is a ring homomorphism from Z % Z into Z % Z. What are the possibilities for f((1, 0))? Determine all ring homomorphisms from Z % Z into Z % Z. In Z, let A 5 2 and B 5 8. Show that the group A/B is isomorphic to the group Z4 but that the ring A/B is not ring-isomorphic to the ring Z4. Let R be a ring with unity and let f be a ring homomorphism from R onto S where S has more than one element. Prove that S has a unity. Show that (Z % Z)/(a % b) is ring-isomorphic to Za % Zb. Determine all ring homomorphisms from Z % Z to Z. Prove that the sum of the squares of three consecutive integers cannot be a square. Let m be a positive integer and let n be an integer obtained from m by rearranging the digits of m in some way. (For example, 72345 is a rearrangement of 35274.) Show that m 2 n is divisible by 9. ping c

17.

a 0

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32. (Test for divisibility by 11) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 11 if and only if a0 2 a1 1 a2 2 ? ? ? (21)kak is divisible by 11. 33. Show that the number 7,176,825,942,116,027,211 is divisible by 9 but not divisible by 11. 34. Show that the number 9,897,654,527,609,805 is divisible by 99. 35. (Test for divisibility by 3) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 3 if and only if ak 1 ak21 1 ? ? ? 1 a1 1 a0 is divisible by 3. 36. (Test for divisibility by 4) Let n be an integer with decimal representation akak21 ? ? ? a1a0. Prove that n is divisible by 4 if and only if a1a0 is divisible by 4. 37. Show that no integer of the form 111,111,111, . . . ,111 is prime. 38. Consider an integer n of the form a, 111,111,111,111,111,111, 111,111,12b. Find values for a and b such that n is divisible by 99. 39. Suppose n is a positive integer written in the form n 5 ak3k 1 ak213k21 1 ? ? ? 1 a13 1 a0, where each of the ai’s is 0, 1, or 2 (the base 3 representative of n). Show that n is even if and only if ak 1 ak21 1 ? ? ? 1 a1 1 a0 is even. 40. Find an analog of the condition given in the previous exercise for characterizing divisibility by 4. 41. In your head, determine (2 ? 1075 1 2)100 mod 3 and (10100 1 1)99 mod 3. 42. Determine all ring homomorphisms from Q to Q. 43. Let R and S be commutative rings with unity. If f is a homomorphism from R onto S and the characteristic of R is nonzero, prove that the characteristic of S divides the characteristic of R. 44. Let R be a commutative ring of prime characteristic p. Show that the Frobenius map x S xp is a ring homomorphism from R to R. 45. Is there a ring homomorphism from the reals to some ring whose kernel is the integers? 46. Show that a homomorphism from a field onto a ring with more than one element must be an isomorphism. 47. Suppose that R and S are commutative rings with unities. Let f be a ring homomorphism from R onto S and let A be an ideal of S. a. If A is prime in S, show that f21(A) 5 {x [ R | f(x) [ A} is prime in R. b. If A is maximal in S, show that f21(A) is maximal in R.

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48. A principal ideal ring is a ring with the property that every ideal has the form a. Show that the homomorphic image of a principal ideal ring is a principal ideal ring. 49. Let R and S be rings. a. Show that the mapping from R % S onto R given by (a, b) S a is a ring homomorphism. b. Show that the mapping from R to R % S given by a S (a, 0) is a one-to-one ring homomorphism. c. Show that R % S is ring-isomorphic to S % R. 50. Show that if m and n are distinct positive integers, then mZ is not ring-isomorphic to nZ. 51. Prove or disprove that the field of real numbers is ring-isomorphic to the field of complex numbers. 52. Show that the only ring automorphism of the real numbers is the identity mapping. 53. Determine all ring homomorphisms from R to R. 54. Suppose that n divides m and that a is an idempotent of Zn (that is, a2 5 a). Show that the mapping x S ax is a ring homomorphism from Zm to Zn. Show that the same correspondence need not yield a ring homomorphism if n does not divide m. 55. Show that the operation of multiplication defined in the proof of Theorem 15.6 is well defined. 56. Let Q["2] 5 {a 1 b"2 | a, b [ Q} and Q["5] 5 {a 1 b"5 | a, b [ Q}. Show that these two rings are not ring-isomorphic. 57. Let Z[i] 5 {a 1 bi | a, b [ Z}. Show that the field of quotients of Z[i] is ring-isomorphic to Q[i] 5 {r 1 si | r, s [ Q}. (This exercise is referred to in Chapter 18.) 58. Let F be a field. Show that the field of quotients of F is ringisomorphic to F. 59. Let D be an integral domain and let F be the field of quotients of D. Show that if E is any field that contains D, then E contains a subfield that is ring-isomorphic to F. (Thus, the field of quotients of an integral domain D is the smallest field containing D.) 60. Explain why a commutative ring with unity that is not an integral domain cannot be contained in a field. (Compare with Theorem 15.6.) 61. Show that the relation ; defined in the proof of Theorem 15.6 is an equivalence relation. 62. Give an example of a ring without unity that is contained in a field. 63. Prove that the set T in the proof of Corollary 3 to Theorem 15.5 is ring-isomorphic to the field of rational numbers.

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64. Suppose that f: R S S is a ring homomorphism and that the image of f is not {0}. If R has a unity and S is an integral domain, show that f carries the unity of R to the unity of S. Give an example to show that the preceding statement need not be true if S is not an integral domain. 65. Let f(x) [ R[x]. If a 1 bi is a complex zero of f(x) (here i 5 "21), show that a 2 bi is a zero of f(x). (This exercise is referred to in Chapter 32.) 66. Let R 5 e c

a b

b d ` a, b [ Z f , and let f be the mapping that a

a b d to a 2 b. b a a. Show that f is a homomorphism. b. Determine the kernel of f. c. Show that R/Ker f is isomorphic to Z. d. Is Ker f a prime ideal? e. Is Ker f a maximal ideal? 67. Show that the prime subfield of a field of characteristic p is ringisomorphic to Zp and that the prime subfield of a field of characteristic 0 is ring-isomorphic to Q. (This exercise is referred to in this chapter.) 68. Let n be a positive integer. Show that there is a ring isomorphism from Z2 to a subring of Z2n if and only if n is odd. 69. Show that Zmn is ring-isomorphic to Zm % Zn when m and n are relatively prime. takes c

Suggested Readings J. A. Gallian and J. Van Buskirk, “The Number of Homomorphisms from Zm into Zn,” American Mathematical Monthly 91 (1984): 196–197. In this article, formulas are given for the number of group homomorphisms from Zm into Zn and the number of ring homomorphisms from Zm into Zn. This article can be downloaded at http://www.d.umn.edu/ ~jgallian/homs.pdf Lillian Kinkade and Joyce Wagner, “When Polynomial Rings Are Principal Ideal Rings,” Journal of Undergraduate Mathematics 23 (1991): 59–62. In this article written by undergraduates, it is shown that R[x] is a principal ideal ring if and only if R < R1 % R2 % ? ? ? % Rn, where each Ri is a field.

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Mohammad Saleh and Hasan Yousef, “The Number of Ring Homomorphisms from Zm1 % ? ? ? % Zmr into Zk1 % ? ? ? % Zk s ,” American Mathematical Monthly 105 (1998): 259–260. This article gives a formula for the number described in the title.

Suggested Website http://www.d.umn.edu/~jgallian/puzzle This site has a math puzzle that is based on the ideas presented in this chapter. The user selects an integer and then proceeds through a series of steps to produce a new integer. Finally, another integer is created by using all but one of the digits of the previous integer in any order. The software then determines the digit not used.

16 Polynomial Rings

Wit lies in recognizing the resemblance among things which differ and the difference between things which are alike. MADAME DE STAEL

Notation and Terminology One of the mathematical concepts that students are most familiar with and most comfortable with is that of a polynomial. In high school, students study polynomials with integer coefficients, rational coefficients, real coefficients, and perhaps even complex coefficients. In earlier chapters of this book, we introduced something that was probably new—polynomials with coefficients from Zn. Notice that all of these sets of polynomials are rings, and, in each case, the set of coefficients is also a ring. In this chapter, we abstract all of these examples into one. Definition Ring of Polynomials over R

Let R be a commutative ring. The set of formal symbols R[x] 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | ai [ R, n is a nonnegative integer} is called the ring of polynomials over R in the indeterminate x. Two elements anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 and bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0 of R[x] are considered equal if and only if ai 5 bi for all nonnegative integers i. (Define ai 5 0 when i . n and bi 5 0 when i . m.)

293

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In this definition, the symbols x, x2, . . . , xn do not represent “unknown” elements or variables from the ring R. Rather, their purpose is to serve as convenient placeholders that separate the ring elements an, an21, . . . , a0. We could have avoided the x’s by defining a polynomial as an infinite sequence a0, a1, a2, . . . , an, 0, 0, 0, . . . , but our method takes advantage of the student’s experience in manipulating polynomials where x does represent a variable. The disadvantage of our method is that one must be careful not to confuse a polynomial with the function determined by a polynomial. For example, in Z3[x], the polynomials f (x) 5 x3 and g(x) 5 x5 determine the same function from Z3 to Z3, since f(a) 5 g(a) for all a in Z3.† But f(x) and g(x) are different elements of Z3[x]. Also, in the ring Zn[x], be careful to reduce only the coefficients and not the exponents modulo n. For example, in Z3[x], 5x 5 2x, but x5 2 x2. To make R[x] into a ring, we define addition and multiplication in the usual way. Definition Addition and Multiplication in R[x]

Let R be a commutative ring and let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 and g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b1x 1 b0 belong to R[x]. Then f(x) 1 g(x) 5 (as 1 bs)xs 1 (as21 1 bs21)xs21 1 ? ? ? 1 (a1 1 b1)x 1 a0 1 b0, where s is the maximum of m and n, ai 5 0 for i . n, and bi 5 0 for i . m. Also, f(x)g(x) 5 cm1nxm1n 1 cm1n21xm1n21 1 ? ? ? 1 c1x 1 c0, where ck 5 akb0 1 ak21b1 1 ? ? ? 1 a1bk21 1 a0bk for k 5 0, . . . , m 1 n.

Although the definition of multiplication might appear complicated, it is just a formalization of the familiar process of using the distributive

†In

general, given f(x) in R[x] and a in R, f(a) means substitute a for x in the formula for f(x). This substitution is a homomorphism from R[x] to R.

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property and collecting like terms. So, just multiply polynomials over a commutative ring R in the same way that polynomials are always multiplied. Here is an example. Consider f(x) 5 2x3 1 x2 1 2x 1 2 and g(x) 5 2x2 1 2x 1 1 in Z3[x]. Then, in our preceding notation, a5 5 0, a4 5 0, a3 5 2, a2 5 1, a1 5 2, a0 5 2, and b5 5 0, b4 5 0, b3 5 0, b2 5 2, b1 5 2, b0 5 1. Now, using the definitions and remembering that addition and multiplication of the coefficients are done modulo 3, we have f(x) 1 g(x) 5 (2 1 0)x3 1 (1 1 2)x2 1 (2 1 2)x 1 (2 1 1) 5 2x3 1 0x2 1 1x 1 0 5 2x3 1 x and f(x) ? g(x) 5 (0 ? 1 1 0 ? 2 1 2 ? 2 1 1 ? 0 1 2 ? 0 1 2 ? 0)x5 1 (0 ? 1 1 2 ? 2 1 1 ? 2 1 2 ? 0 1 2 ? 0)x4 1 (2 ? 1 1 1 ? 2 1 2 ? 2 1 2 ? 0)x3 1 (1 ? 1 1 2 ? 2 1 2 ? 2)x2 1 (2 ? 1 1 2 ? 2)x 1 2 ? 1 5 x5 1 0x4 1 2x3 1 0x2 1 0x 1 2 5 x5 1 2x3 1 2 Our definitions for addition and multiplication of polynomials were formulated so that they are commutative and associative, and so that multiplication is distributive over addition. We leave the verification that R[x] is a ring to the reader. It is time to introduce some terminology for polynomials. If f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0, where an 2 0, we say that f(x) has degree n; the term an is called the leading coefficient of f(x), and if the leading coefficient is the multiplicative identity element of R, we say that f(x) is a monic polynomial. The polynomial f(x) 5 0 has no degree. Polynomials of the form f(x) 5 a0 are called constant. We often write deg f(x) 5 n to indicate that f(x) has degree n. In keeping with our experience with polynomials with real coefficients, we adopt the following notational conventions: We may insert or delete terms of the form 0xk; 1xk will be denoted by xk; 1 (2ak)xk will be denoted by 2akxk. Very often properties of R carry over to R[x]. Our first theorem is a case in point.

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Theorem 16.1 D an Integral Domain Implies D[x] an Integral Domain If D is an integral domain, then D [x] is an integral domain.

PROOF Since we already know that D[x] is a ring, all we need to show is that D[x] is commutative with a unity and has no zero-divisors. Clearly, D[x] is commutative whenever D is. If 1 is the unity element of D, it is obvious that f(x) 5 1 is the unity element of D[x]. Finally, suppose that f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 and g(x) 5 bmxm 1 bm21xm21 1 ? ? ? 1 b0, where an 2 0 and bm 2 0. Then, by definition, f(x)g(x) has leading coefficient anbm and, since D is an integral domain, anbm 2 0.

The Division Algorithm and Consequences One of the properties of integers that we have used repeatedly is the division algorithm: If a and b are integers and b 2 0, then there exist unique integers q and r such that a 5 bq 1 r, where 0 # r , |b|. The next theorem is the analogous statement for polynomials over a field. Theorem 16.2 Division Algorithm for F[x] Let F be a field and let f(x) and g(x) [ F[x] with g(x) 2 0. Then there exist unique polynomials q(x) and r(x) in F[x] such that f(x) 5 g(x)q(x) 1 r(x) and either r(x) 5 0 or deg r(x) , deg g(x).

PROOF We begin by showing the existence of q(x) and r(x). If f(x) 5 0 or deg f(x) , deg g(x), we simply set q(x) 5 0 and r(x) 5 f(x). So, we may assume that n 5 deg f(x) $ deg g(x) 5 m and let f(x) 5 anxn 1 ? ? ? 1 a0 and g(x) 5 bmxm 1 ? ? ? 1 b0. The idea behind this proof is to begin just as if you were going to “long divide” g(x) into f(x), then use the Second Principle of Mathematical Induction on deg f(x) to finish up. Thus, resorting to long division, we let f1(x) 5

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f(x) 2 anbm21xn2mg(x).† Then, f1(x) 5 0 or deg f1(x) , deg f(x); so, by our induction hypothesis, there exist q1(x) and r1(x) in F[x] such that f1(x) 5 g(x)q1(x) 1 r1(x), where r1(x) 5 0 or deg r1(x) , deg g(x). [Technically, we should get the induction started by proving the case in which deg f(x) 5 0, but this is trivial.] Thus, f(x) 5 anbm21xn2mg(x) 1 f1(x) 5 anbm21xn2mg(x) 1 q1(x)g(x) 1 r1(x) 5 [anbm21xn2m 1 q1(x)]g(x) 1 r1(x). So, the polynomials q(x) 5 anbm21xn2m 1 q1(x) and r(x) 5 r1(x) have the desired properties. To prove uniqueness, suppose that f(x) 5 g(x)q(x) 1 r(x) and f(x) 5 g(x) q(x) 1 r(x), where r(x) 5 0 or deg r(x) , deg g(x) and r(x) 5 0 or deg r (x) , deg g(x). Then, subtracting these two equations, we obtain 0 5 g(x)[q(x) 2 q(x)] 1 [r(x) 2 r(x)] or r(x) 2 r(x) 5 g(x)[q(x) 2 q(x)]. Thus, r(x) 2 r(x) is 0, or the degree of r (x) 2 r(x) is at least that of g(x). Since the latter is clearly impossible, we have r (x) 5 r(x) and q(x) 5 q (x) as well. The polynomials q(x) and r(x) in the division algorithm are called the quotient and remainder in the division of f(x) by g(x). When the ring of coefficients of a polynomial ring is a field, we can use the long division process to determine the quotient and remainder.

†For

example,

(3>2)x2

2x2 1 2 q3x4

1x11 3x4 1 3x2 2 3x2 1 x 1 1

So, 23x2 1 x 1 1 5 3x4 1 x 1 1 2 (3/2)x2(2x2 1 2) In general, anbm21xn2m . . . bm x 1 qan xn 1 . . . a xn 1 . . . m

n

f1 (x)

So, f1(x) 5 (anxn 1 ? ? ?) 2 anbm21xn2m(bmxm 1 ? ? ?)

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EXAMPLE 1 To find the quotient and remainder upon dividing f(x) 5 3x4 1 x3 1 2x2 1 1 by g(x) 5 x2 1 4x 1 2, where f(x) and g(x) belong to Z5[x], we may proceed by long division, provided we keep in mind that addition and multiplication are done modulo 5. Thus, 3x2 1 4x x2 1 4x 1 2 q3x4 1 x3 1 2x2 11 4 3 2 3x 1 2x 1 x 4x3 1 x2 11 3 2 4x 1 x 1 3x 2x 1 1 So, 3x2 1 4x is the quotient and 2x 1 1 is the remainder. Therefore, 3x4 1 x3 1 2x2 1 1 5 (x2 1 4x 1 2)(3x2 1 4x) 1 2x 1 1. Let D be an integral domain. If f(x) and g(x) [ D[x], we say that g(x) divides f(x) in D[x] [and write g(x) | f(x)] if there exists an h(x) [ D[x] such that f(x) 5 g(x)h(x). In this case, we also call g(x) a factor of f(x). An element a is a zero (or a root) of a polynomial f(x) if f(a) 5 0. [Recall that f(a) means substitute a for x in the expression for f(x).] When F is a field, a [ F, and f(x) [ F[x], we say that a is a zero of multiplicity k (k $ 1) if (x 2 a)k is a factor of f(x) but (x 2 a)k11 is not a factor of f(x). With these definitions, we may now give several important corollaries of the division algorithm. No doubt you have seen these for the special case where F is the field of real numbers. Corollary 1 The Remainder Theorem Let F be a field, a [ F, and f(x) [ F[x]. Then f(a) is the remainder in the division of f(x) by x 2 a.

PROOF The proof of Corollary 1 is left as an exercise (Exercise 5). Corollary 2 The Factor Theorem Let F be a field, a [ F, and f(x) [ F[x]. Then a is a zero of f(x) if and only if x 2 a is a factor of f(x).

PROOF The proof of Corollary 2 is left as an exercise (Exercise 7).

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Corollary 3 Polynomials of Degree n Have at Most n Zeros A polynomial of degree n over a field has at most n zeros, counting multiplicity.

PROOF We proceed by induction on n. Clearly, a polynomial of degree 0 over a field has no zeros. Now suppose that f(x) is a polynomial of degree n over a field and a is a zero of f(x) of multiplicity k. Then, f(x) 5 (x 2 a)kq(x) and q(a) 2 0; and, since n 5 deg f(x) 5 deg (x 2 a)k q(x) 5 k 1 deg q(x), we have k # n (see Exercise 17). If f(x) has no zeros other than a, we are done. On the other hand, if b 2 a and b is a zero of f(x), then 0 5 f(b) 5 (b 2 a)kq(b), so that b is also a zero of q(x) with the same multiplicity as it has for f (x) (see Exercise 19). By the Second Principle of Mathematical Induction, we know that q(x) has at most deg q(x) 5 n 2 k zeros, counting multiplicity. Thus, f(x) has at most k 1 n 2 k 5 n zeros, counting multiplicity. We remark that Corollary 3 is not true for arbitrary polynomial rings. For example, the polynomial x2 1 3x 1 2 has four zeros in Z6. (See Exercise 3.) Lagrange was the first to prove Corollary 3 for polynomials in Zp[x]. EXAMPLE 2 The Complex Zeros of xn 2 1 We find all complex zeros of xn 2 1. Let v 5 cos(360°/n) 1 i sin(360°/n). It follows from DeMoivre’s Theorem (see Example 7 in Chapter 0) that vn 5 1 and vk 2 1 for 1 # k , n. Thus, each of 1, v, v2, . . . , vn21 is a zero of xn 2 1 and, by Corollary 3, there are no others. The complex number v in Example 2 is called the primitive nth root of unity. We conclude this chapter with an important theoretical application of the division algorithm, but first an important definition. Definition Principal Ideal Domain (PID)

A principal ideal domain is an integral domain R in which every ideal has the form a 5 {ra | r [ R} for some a in R.

Theorem 16.3 F[x] Is a PID Let F be a field. Then F[x] is a principal ideal domain.

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PROOF By Theorem 16.1, we know that F[x] is an integral domain. Now, let I be an ideal in F[x]. If I 5 {0}, then I 5 0. If I 2 {0}, then among all the elements of I, let g(x) be one of minimum degree. We will show that I 5 g(x). Since g(x) [ I, we have g(x) # I. Now let f(x) [ I. Then, by the division algorithm, we may write f(x) 5 g(x)q(x) 1 r(x), where r(x) 5 0 or deg r(x) , deg g(x). Since r(x) 5 f(x) 2 g(x)q(x) [ I, the minimality of deg g(x) implies that the latter condition cannot hold. So, r(x) 5 0 and, therefore, f(x) [ g(x). This shows that I # g(x). The proof of Theorem 16.3 also establishes the following. Theorem 16.4 Criterion for I 5 g(x) Let F be a field, I a nonzero ideal in F[x], and g(x) an element of F[x]. Then, I 5 8 g(x)9 if and only if g(x) is a nonzero polynomial of minimum degree in I.

As an application of the First Isomorphism Theorem for Rings (Theorem 15.3) and Theorem 16.4, we verify the remark we made in Example 12 in Chapter 14 that the ring R[x]/x2 1 1 is isomorphic to the ring of complex numbers. EXAMPLE 3 Consider the homomorphism f from R[x] onto C given by f(x) → f(i) (that is, evaluate a polynomial in R[x] at i). Then x2 1 1 [ Ker f and is clearly a polynomial of minimum degree in Ker f. Thus, Ker f 5 x2 1 1 and R[x]/x2 1 1 is isomorphic to C.

Exercises If I feel unhappy, I do mathematics to become happy. If I am happy, I do mathematics to keep happy. PAUL TURÁN

1. Let f(x) 5 4x3 1 2x2 1 x 1 3 and g(x) 5 3x4 1 3x3 1 3x2 1 x 1 4, where f(x), g(x) [ Z5[x]. Compute f(x) 1 g(x) and f(x) ? g(x). 2. In Z3[x], show that the distinct polynomials x4 1 x and x2 1 x determine the same function from Z3 to Z3. 3. Show that x2 1 3x 1 2 has four zeros in Z6. (This exercise is referred to in this chapter.)

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4. If R is a commutative ring, show that the characteristic of R[x] is the same as the characteristic of R. 5. Prove Corollary 1 of Theorem 16.2. 6. List all the polynomials of degree 2 in Z2[x]. 7. Prove Corollary 2 of Theorem 16.2. 8. Let R be a commutative ring. Show that R[x] has a subring isomorphic to R. 9. If f: R → S is a ring homomorphism, define f:R[x] → S[x] by (anxn 1 ? ? ? 1 a0) → f(an)xn 1 ? ? ? 1 f(a0). Show that f is a ring homomorphism. (This exercise is referred to in Chapter 33.) 10. If the rings R and S are isomorphic, show that R[x] and S[x] are isomorphic. 11. Let f(x) 5 x3 1 2x 1 4 and g(x) 5 3x 1 2 in Z5[x]. Determine the quotient and remainder upon dividing f(x) by g(x). 12. Let f(x) 5 5x4 1 3x3 1 1 and g(x) 5 3x2 1 2x 1 1 in Z7[x]. Determine the quotient and remainder upon dividing f(x) by g(x). 13. Show that the polynomial 2x 1 1 in Z4[x] has a multiplicative inverse in Z4[x]. 14. Are there any nonconstant polynomials in Z[x] that have multiplicative inverses? Explain your answer. 15. Let p be a prime. Are there any nonconstant polynomials in Zp[x] that have multiplicative inverses? Explain your answer. 16. Show that Corollary 3 of Theorem 16.2 is false for any commutative ring that has a zero divisor. 17. (Degree Rule) Let D be an integral domain and f(x), g(x) [ D[x]. Prove that deg ( f(x) ? g(x)) 5 deg f(x) 1 deg g(x). Show, by example, that for commutative ring R it is possible that deg f(x)g(x) , deg f(x) 1 deg g(x) where f(x) and g(x) are nonzero elements in R[x]. (This exercise is referred to in this chapter, Chapter 17, and Chapter 18.) 18. Prove that the ideal x in Q[x] is maximal. 19. Let f(x) belong to F[x], where F is a field. Let a be a zero of f(x) of multiplicity n, and write f(x) 5 (x 2 a)nq(x). If b ? a is a zero of q(x), show that b has the same multiplicity as a zero of q(x) as it does for f(x). (This exercise is referred to in this chapter.) 20. Prove that for any positive integer n, a field F can have at most a finite number of elements of order at most n. 21. Let F be an infinite field and let f(x) [ F[x]. If f(a) 5 0 for infinitely many elements a of F, show that f(x) 5 0.

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22. Let F be an infinite field and let f(x), g(x) [ F[x]. If f(a) 5 g(a) for infinitely many elements a of F, show that f(x) 5 g(x). 23. Let F be a field and let p(x) [ F[x]. If f(x), g(x) [ F[x] and deg f(x) , deg p(x) and deg g(x) , deg p(x), show that f(x) 1 p(x) 5 g(x) 1 p(x) implies f(x) 5 g(x). (This exercise is referred to in Chapter 20.) 24. Prove that Z[x] is not a principal ideal domain. (Compare this with Theorem 16.3.) 25. Find a polynomial with integer coefficients that has 1/2 and 21/3 as zeros. 26. Let f(x) [ R[x]. Suppose that f(a) 5 0 but f9(a) 2 0, where f9(x) is the derivative of f(x). Show that a is a zero of f(x) of multiplicity 1. 27. Show that Corollary 2 of Theorem 16.2 is true over any commutative ring with unity. 28. Show that Corollary 3 of Theorem 16.2 is true for polynomials over integral domains. 29. Let F be a field and let I 5 {anxn 1 an21xn21 1 ? ? ? 1 a0 | an, an21, . . . , a0 [ F and an 1 an21 1 ? ? ? 1 a0 5 0}. Show that I is an ideal of F[x] and find a generator for I. 30. Let F be a field and let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ F[x]. Prove that x 2 1 is a factor of f(x) if and only if an 1 an21 1 ? ? ? 1 a0 5 0. 31. Let m be a fixed positive integer. For any integer a, let a denote a mod m. Show that the mapping of f: Z[x] → Zm[x] given by f(anxn 1 an21xn21 1 ? ? ? 1 a0) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 is a ring homomorphism. (This exercise is referred to in Chapters 17 and 33.) 32. Find infinitely many polynomials f(x) in Z3[x] such that f(a) 5 0 for all a in Z3. 33. For every prime p, show that xp21 2 1 5 (x 2 1)(x 2 2) ? ? ? [x 2 ( p 2 1)] 34. 35. 36. 37.

in Zp[x]. (Wilson’s Theorem) For every integer n . 1, prove that (n 2 1)! mod n 5 n 2 1 if and only if n is prime. For every prime p, show that ( p 2 2)! mod p 5 1. Find the remainder upon dividing 98! by 101. Prove that (50!)2 mod 101 5 21 mod 101.

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38. If I is an ideal of a ring R, prove that I[x] is an ideal of R[x]. 39. Give an example of a commutative ring R with unity and a maximal ideal I of R such that I[x] is not a maximal ideal of R[x]. 40. Let R be a commutative ring with unity. If I is a prime ideal of R, prove that I[x] is a prime ideal of R[x]. 41. Let F be a field, and let f(x) and g(x) belong to F[x]. If there is no polynomial of positive degree in F[x] that divides both f(x) and g(x) [in this case, f(x) and g(x) are said to be relatively prime], prove that there exist polynomials h(x) and k(x) in F[x] with the property that f(x)h(x) 1 g(x)k(x) 5 1. (This exercise is referred to in Chapter 20.) 42. Prove that Q[x]/x2 2 2 is ring-isomorphic to Q["2 ] 5 {a 1 b"2 | a, b [ Q}. 43. Let f(x) [ R[x]. If f(a) 5 0 and f 9(a) 5 0 [f 9(a) is the derivative of f(x) at a], show that (x 2 a)2 divides f(x). 44. Let F be a field and let I 5 {f(x) [ F[x] | f(a) 5 0 for all a in F}. Prove that I is an ideal in F[x]. Prove that I is infinite when F is finite and I 5 {0} when F is infinite. When F is finite, find a monic polynomial g(x) such that I 5 g(x). 45. Let g(x) and h(x) belong to Z[x] and let h(x) be monic. If h(x) divides g(x) in Q[x], show that h(x) divides g(x) in Z[x]. (This exercise is referred to in Chapter 33.) 46. For any field F, recall that F(x) denotes the field of quotients of the ring F[x]. Prove that there is no element in F(x) whose square is x. 47. Let F be a field. Show that there exist a, b [ F with the property that x2 1 x 1 1 divides x43 1 ax 1 b. 48. Let f(x) 5 am x m 1 am21x m21 1 ? ? ? 1 a0 and g(x) 5 bn xn 1 bn21x n21 1 ? ? ? 1 b0 belong to Q[x] and suppose that f 8 g belongs to Z[x]. Prove that aibj is an integer for every i and j. 49. Let f (x) belong to Z[x]. If a mod m 5 b mod m, prove that f (a) mod m 5 f(b) mod m. Prove that if both f(0) and f(1) are odd then f has no zero in Z. 50. Find the remainder when x51 is divided by x 1 4 in Z7[x]. 51. Show that 1 is the only solution of x25 2 1 5 0 in Z37.

Saunders Mac Lane The 1986 Steele Prize for cumulative influence is awarded to Saunders Mac Lane for his many contributions to algebra and algebraic topology, and in particular for his pioneering work in homological and categorical algebra. Citation for the Steele Prize

SAUNDERS MAC LANE ranks among the most influential mathematicians in the twentieth century. He was born on August 4, 1909, in Norwich, Connecticut. In 1933, at the height of the Depression, he was newly married; despite having degrees from Yale, the University of Chicago, and the University of Göttingen, he had no prospects for a position at a college or university. After applying for employment as a master at a private preparatory school for boys, Mac Lane received a two-year instructorship at Harvard in 1934. He then spent a year at Cornell and a year at the University of Chicago before returning to Harvard in 1938. In 1947, he went back to Chicago permanently. Much of Mac Lane’s work focuses on the interconnections among algebra, topology,

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and geometry. His book, Survey of Modern Algebra, coauthored with Garrett Birkhoff, influenced generations of mathematicians and is now a classic. Mac Lane has served as president of the Mathematical Association of America and the American Mathematical Society. He was elected to the National Academy of Sciences, received the National Medal of Science and supervised 41 Ph.D. theses. Mac Lane died April 14, 2005, at age of 95. To find more information about Mac Lane, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Factorization 17 of Polynomials The value of a principle is the number of things it will explain. RALPH WALDO EMERSON

Reducibility Tests In high school, students spend much time factoring polynomials and finding their zeros. In this chapter, we consider the same problems in a more abstract setting. To discuss factorization of polynomials, we must first introduce the polynomial analog of a prime integer. Definition Irreducible Polynomial, Reducible Polynomial

Let D be an integral domain. A polynomial f(x) from D[x] that is neither the zero polynomial nor a unit in D[x] is said to be irreducible over D if, whenever f(x) is expressed as a product f(x) 5 g(x)h(x), with g(x) and h(x) from D[x], then g(x) or h(x) is a unit in D[x]. A nonzero, nonunit element of D[x] that is not irreducible over D is called reducible over D.

In the case that an integral domain is a field F, it is equivalent and more convenient to define a nonconstant f(x) [ F[x] to be irreducible if f(x) cannot be expressed as a product of two polynomials of lower degree. EXAMPLE 1 The polynomial f(x) 5 2x2 1 4 is irreducible over Q but reducible over Z, since 2x2 1 4 5 2(x2 1 2) and neither 2 nor x2 1 2 is a unit in Z[x]. EXAMPLE 2 The polynomial f(x) 5 2x2 1 4 is irreducible over R but reducible over C. EXAMPLE 3 The polynomial x2 2 2 is irreducible over Q but reducible over R.

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EXAMPLE 4 The polynomial x2 1 1 is irreducible over Z3 but reducible over Z5. In general, it is a difficult problem to decide whether or not a particular polynomial is reducible over an integral domain, but there are special cases when it is easy. Our first theorem is a case in point. It applies to the three preceding examples. Theorem 17.1 Reducibility Test for Degrees 2 and 3 Let F be a field. If f(x) [ F[x] and deg f(x) is 2 or 3, then f(x) is reducible over F if and only if f(x) has a zero in F.

PROOF Suppose that f(x) 5 g(x)h(x), where both g(x) and h(x) belong to F[x] and have degrees less than that of f(x). Since deg f(x) 5 deg g(x) 1 deg h(x) (Exercise 17 in Chapter 16) and deg f(x) is 2 or 3, at least one of g(x) and h(x) has degree 1. Say g(x) 5 ax 1 b. Then, clearly, 2a21b is a zero of g(x) and therefore a zero of f(x) as well. Conversely, suppose that f(a) 5 0, where a [ F. Then, by the Factor Theorem, we know that x 2 a is a factor of f(x) and, therefore, f(x) is reducible over F. Theorem 17.1 is particularly easy to use when the field is Zp, because, in this case, we can check for reducibility of f(x) by simply testing to see if f(a) 5 0 for a 5 0, 1, . . . , p 2 1. For example, since 2 is a zero of x2 1 1 over Z5 , x2 1 1 is reducible over Z5 . On the other hand, because neither 0, 1, nor 2 is a zero of x2 1 1 over Z3, x2 1 1 is irreducible over Z3. Note that polynomials of degree larger than 3 may be reducible over a field, even though they do not have zeros in the field. For example, in Q[x], the polynomial x4 1 2x2 1 1 is equal to (x2 1 1)2, but has no zeros in Q. Our next three tests deal with polynomials with integer coefficients. To simplify the proof of the first of these, we introduce some terminology and isolate a portion of the argument in the form of a lemma. Definition Content of Polynomial, Primitive Polynomial

The content of a nonzero polynomial anxn 1 an21xn21 1 ? ? ? 1 a0, where the a’s are integers, is the greatest common divisor of the integers an, an21, . . . , a0. A primitive polynomial is an element of Z[x] with content 1.

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Gauss’s Lemma The product of two primitive polynomials is primitive.

PROOF Let f(x) and g(x) be primitive polynomials, and suppose that f(x)g(x) is not primitive. Let p be a prime divisor of the content of f(x)g(x), and let f (x), g(x), and f(x)g(x) be the polynomials obtained from f(x), g(x), and f(x)g(x) by reducing the coefficients modulo p. Then, f (x) and g(x) belong to the integral domain Zp[x] and f (x)g(x) 5 f(x)g(x) 5 0, the zero element of Zp[x] (see Exercise 31 in Chapter 16). Thus, f (x) 5 0 or g(x) 5 0. This means that either p divides every coefficient of f(x) or p divides every coefficient of g(x). Hence, either f(x) is not primitive or g(x) is not primitive. This contradiction completes the proof. Remember that the question of reducibility depends on which ring of coefficients one permits. Thus, x2 2 2 is irreducible over Z but reducible over Q["2]. In Chapter 20, we will prove that every polynomial of degree greater than 1 with coefficients from an integral domain is reducible over some field. Theorem 17.2 shows that in the case of polynomials irreducible over Z, this field must be larger than the field of rational numbers. Theorem 17.2 Reducibility over Q Implies Reducibility Over Z Let f(x) [ Z[x]. If f(x) is reducible over Q, then it is reducible over Z.

PROOF Suppose that f(x) 5 g(x)h(x), where g(x) and h(x) [ Q[x]. Clearly, we may assume that f(x) is primitive because we can divide both f(x) and g(x) by the content of f(x). Let a be the least common multiple of the denominators of the coefficients of g(x), and b the least common multiple of the denominators of the coefficients of h(x). Then abf(x) 5 ag(x) ? bh(x), where ag(x) and bh(x) [ Z[x]. Let c1 be the content of ag(x) and let c2 be the content of bh(x). Then ag(x) 5 c1g1(x) and bh(x) 5 c2h1(x), where both g1(x) and h1(x) are primitive and abf(x) 5 c1c2g1(x)h1(x). Since f(x) is primitive, the content of abf(x) is ab. Also, since the product of two primitive polynomials is primitive, it follows that the content of c1c2g1(x)h1(x) is c1c2. Thus, ab 5 c1c2 and f(x) 5 g1(x)h1(x), where g1(x) and h1(x) [ Z[x] and deg g1(x) 5 deg g(x) and deg h1(x) 5 deg h(x).

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EXAMPLE 5 We illustrate the proof of Theorem 17.2 by tracing through it for the polynomial f(x) 5 6x2 1 x 2 2 5 (3x 2 3/2)(2x 1 4/3) 5 g(x)h(x). In this case we have a 5 2, b 5 3, c1 5 3, c2 5 2, g1(x) 5 2x 2 1, and h1(x) 5 3x 1 2, so that 2 ? 3(6x2 1 x 2 2) 5 3 ? 2(2x 2 1)(3x 1 2) or 6x2 1 x 2 2 5 (2x 2 1)(3x 1 2).

Irreducibility Tests Theorem 17.1 reduces the question of irreducibility of a polynomial of degree 2 or 3 to one of finding a zero. The next theorem often allows us to simplify the problem even further. Theorem 17.3 Mod p Irreducibility Test Let p be a prime and suppose that f(x) [ Z[x] with deg f(x) $ 1. Let f (x) be the polynomial in Zp[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f (x) is irreducible over Zp and deg f (x) 5 deg f(x), then f(x) is irreducible over Q.

PROOF It follows from the proof of Theorem 17.2 that if f(x) is reducible over Q, then f(x) 5 g(x)h(x) with g(x), h(x) [ Z[x], and both g(x) and h(x) have degree less than that of f(x). Let f (x) , g(x), and h(x) be the polynomials obtained from f(x), g(x), and h(x) by reducing all the coefficients modulo p. Since deg f(x) 5 deg f (x) , we have deg g(x) # deg g(x) , deg f (x) and deg h(x) # deg h(x) , deg f (x) . But, f (x) 5 g(x)h(x), and this contradicts our assumption that f (x) is irreducible over Zp. EXAMPLE 6 Let f(x) 5 21x3 2 3x2 1 2x 1 9. Then, over Z2, we have f (x) 5 x3 1 x2 1 1 and, since f (0) 5 1 and f (1) 5 1, we see that f (x) is irreducible over Z2. Thus, f(x) is irreducible over Q. Notice that, over Z3, f (x) 5 2x is irreducible, but we may not apply Theorem 17.3 to conclude that f(x) is irreducible over Q. Be careful not to use the converse of Theorem 17.3. If f(x) [ Z[x] and f (x) is reducible over Zp for some p, f(x) may still be irreducible over Q. For example, consider f(x) 5 21x3 2 3x2 1 2x 1 8. Then, over Z2, f (x) 5 x3 1 x2 5 x2(x 1 1). But over Z5, f (x) has no zeros and therefore is irreducible over Z5. So, f(x) is irreducible over Q. Note that this example shows that the Mod p Irreducibility Test may fail for some p and work for others. To conclude that a particular f(x) in Z[x] is

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irreducible over Q, all we need to do is find a single p for which the corresponding polynomial f (x) in Zp is irreducible. However, this is not always possible, since f(x) 5 x4 1 1 is irreducible over Q but reducible over Zp for every prime p. (See Exercise 29.) The Mod p Irreducibility Test can also be helpful in checking for irreducibility of polynomials of degree greater than 3 and polynomials with rational coefficients. EXAMPLE 7 Let f(x) 5 (3/7)x4 2 (2/7)x2 1 (9/35)x 1 3/5. We will show that f(x) is irreducible over Q. First, let h(x) 5 35f(x) 5 15x4 2 10x2 1 9x 1 21. Then f(x) is irreducible over Q if h(x) is irreducible over Z. Next, applying the Mod 2 Irreducibility Test to h(x), we get h(x) 5 x4 1 x 1 1. Clearly, h(x) has no zeros in Z2. Furthermore, h(x) has no quadratic factor in Z2[x] either. [For if so, the factor would have to be either x2 1 x 1 1 or x2 1 1. Long division shows that x2 1 x 1 1 is not a factor, and x2 1 1 cannot be a factor because it has a zero whereas h(x) does not.] Thus h(x) is irreducible over Z2[x]. This guarantees that h(x) is irreducible over Q. EXAMPLE 8 Let f(x) 5 x 5 1 2x 1 4. Obviously, neither Theorem 17.1 nor the Mod 2 Irreducibility Test helps here. Let’s try mod 3. Substitution of 0, 1, and 2 into f (x) does not yield 0, so there are no linear factors. But f (x) may have a quadratic factor. If so, we may assume it has the form x2 1 ax 1 b (see Exercise 5). This gives nine possibilities to check. We can immediately rule out each of the nine that has a zero over Z3, since f (x) does not have one. This leaves only x2 1 1, x2 1 x 1 2, and x2 1 2x 1 2 to check. These are eliminated by long division. So, since f (x) is irreducible over Z3, f(x) is irreducible over Q. (Why is it unnecessary to check for cubic or fourth-degree factors?) Another important irreducibility test is the following one, credited to Ferdinand Eisenstein (1823–1852), a student of Gauss. The corollary was first proved by Gauss by a different method. Theorem 17.4 Eisenstein’s Criterion (1850) Let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]. If there is a prime p such that p B an, p | an21, . . . , p | a0 and p2 B a0, then f(x) is irreducible over Q.

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PROOF If f(x) is reducible over Q, we know by Theorem 17.2 that there exist elements g(x) and h(x) in Z[x] such that f(x) 5 g(x)h(x) and 1 # deg g(x), and 1 # deg h(x) , n. Say g(x) 5 br xr 1 ? ? ? 1 b0 and h(x) 5 cs xs 1 ? ? ? 1 c0. Then, since p | a0, p2 B a0, and a0 5 b0c0, it follows that p divides one of b0 and c0 but not the other. Let us say p | b0 and p B c0. Also, since p B an 5 brcs, we know that p B br. So, there is a least integer t such that p B bt. Now, consider at 5 btc0 1 bt21c1 1 ? ? ? 1 b0ct. By assumption, p divides at and, by choice of t, every summand on the right after the first one is divisible by p. Clearly, this forces p to divide btc0 as well. This is impossible, however, since p is prime and p divides neither bt nor c0. Corollary Irreducibility of pth Cyclotomic Polynomial For any prime p, the pth cyclotomic polynomial Fp(x) 5

xp 2 1 5 xp21 1 xp22 1 ? ? ? 1 x 1 1 x21

is irreducible over Q.

PROOF Let f(x)5Fp(x11)5

(x11) p 21 p p p 5x p21 1 a b x p22 1 a b x p23 1 . . . 1 a b. (x11)21 1 2 1

Then, since every coefficient except that of xp21 is divisible by p and the constant term is not divisible by p2, by Eisenstein’s Criterion, f(x) is irreducible over Q. So, if Fp(x) 5 g(x)h(x) were a nontrivial factorization of Fp(x) over Q, then f(x) 5 Fp(x 1 1) 5 g(x 1 1) ? h(x 1 1) would be a nontrivial factorization of f(x) over Q. Since this is impossible, we conclude that Fp(x) is irreducible over Q. EXAMPLE 9 The polynomial 3x5 1 15x4 2 20x3 1 10x 1 20 is irreducible over Q because 5 B 3 and 25 B 20 but 5 does divide 15, 220, 10, and 20. The principal reason for our interest in irreducible polynomials stems from the fact that there is an intimate connection among them, maximal ideals, and fields. This connection is revealed in the next theorem and its first corollary.

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Theorem 17.5  p(x) Is Maximal If and Only If p(x) Is Irreducible Let F be a field and let p(x) [ F[x]. Then p(x) is a maximal ideal in F[x] if and only if p(x) is irreducible over F.

PROOF Suppose first that p(x) is a maximal ideal in F[x]. Clearly, p(x) is neither the zero polynomial nor a unit in F[x], because neither {0} nor F[x] is a maximal ideal in F[x]. If p(x) 5 g(x)h(x) is a factorization of p(x) over F, then p(x) # g(x) # F[x]. Thus, p(x) 5 g(x) or F[x] 5 g(x). In the first case, we must have deg p(x) 5 deg g(x). In the second case, it follows that deg g(x) 5 0 and, consequently, deg h(x) 5 deg p(x). Thus, p(x) cannot be written as a product of two polynomials in F[x] of lower degree. Now, suppose that p(x) is irreducible over F. Let I be any ideal of F[x] such that  p(x) # I # F[x]. Because F[x] is a principal ideal domain, we know that I 5 g(x) for some g(x) in F[x]. So, p(x) [ g(x) and, therefore, p(x) 5 g(x)h(x), where h(x) [ F[x]. Since p(x) is irreducible over F, it follows that either g(x) is a constant or h(x) is a constant. In the first case, we have I 5 F[x]; in the second case, we have p(x) 5 g(x) 5 I. So, p(x) is maximal in F[x]. Corollary 1 F[x]/ p(x) Is a Field Let F be a field and p(x) an irreducible polynomial over F. Then F[x]/p(x) is a field.

PROOF This follows directly from Theorems 17.5 and 14.4. The next corollary is a polynomial analog of Euclid’s Lemma for primes (see Chapter 0). Corollary 2 p(x) | a(x)b(x) Implies p(x) | a(x) or p(x) | b(x) Let F be a field and let p(x), a(x), b(x) [ F[x]. If p(x) is irreducible over F and p(x) | a(x)b(x), then p(x) | a(x) or p(x) | b(x).

PROOF Since p(x) is irreducible, F[x]/ p(x) is a field and, therefore, an integral domain. From Theorem 14.3, we know that p(x) is a prime ideal, and since p(x) divides a(x)b(x), we have a(x)b(x) [ p(x). Thus, a(x) [ p(x) or b(x) [ p(x). This means that p(x) | a(x) or p(x) | b(x). The next two examples put the theory to work.

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EXAMPLE 10 We construct a field with eight elements. By Theorem 17.1 and Corollary 1 of Theorem 17.5, it suffices to find a cubic polynomial over Z2 that has no zero in Z2. By inspection, x3 1 x 1 1 fills the bill. Thus, Z2[x]/x3 1 x 1 1 5 {ax2 1 bx 1 c 1 x3 1 x 1 1 | a, b, c [ Z2} is a field with eight elements. For practice, let us do a few calculations in this field. Since the sum of two polynomials of the form ax2 1 bx 1 c is another one of the same form, addition is easy. For example, (x2 1 x 1 1 1 x3 1 x 1 1) 1 (x2 1 1 1 x3 1 x 1 1) 5 x 1 x3 1 x 1 1. On the other hand, multiplication of two coset representatives need not yield one of the original eight coset representatives: (x2 1 x 1 1 1 x3 1 x 1 1) ? (x2 1 1 1 x3 1 x 1 1) 5 x4 1 x3 1 x 1 1 1 x3 1 x 1 1 5 x4 1 x3 1 x 1 1 (since the ideal absorbs the last three terms). How do we express this in the form ax2 1 bx 1 c 1 x3 1 x 1 1? One way is to long divide x4 by x3 1 x 1 1 to obtain the remainder of x2 1 x (just as one reduces 12 1 5 to 2 1 5 by dividing 12 by 5 to obtain the remainder 2). Another way is to observe that x3 1 x 1 1 1 x3 1 x 1 1 5 0 1 x3 1 x 1 1 implies x3 1 x3 1 x 1 1 5 x 1 1 1 x3 1 x 1 1. Thus, we may multiply both sides by x to obtain x4 1 x3 1 x 1 1 5 x2 1 x 1 x3 1 x 1 1. Similarly, (x2 1 x 1 kx3 1 x 1 1l) ? (x 1 kx3 1 x 1 1l) 5 x3 1 x2 1 kx3 1 x 1 1l 5 x2 1 x 1 1 1 kx3 1 x 1 1l. A partial multiplication table for this field is given in Table 17.1. To simplify the notation, we indicate a coset by its representative only. Table 17.1 A Partial Multiplication Table for Example 10

1 x x11 x2 x2 1 1

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x2 x11 x2 1 x 1 1 x2 1 x x

x2 1 1 1 x2 x x2 1 x 1 1

x2 1 x x2 1 x 1 1 1 x2 1 1 x11

x2 1 x 1 1 x2 1 1 x 1 x2 1 x

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(Complete the table yourself. Keep in mind that x3 can be replaced by x 1 1 and x4 by x2 1 x.) EXAMPLE 11 Since x2 1 1 has no zero in Z3, it is irreducible over Z3. Thus, Z3[x]/x2 1 1 is a field. Analogous to Example 12 in Chapter 14, Z3[x]/x2 1 1 5 {ax 1 b 1 x2 1 1 | a, b [ Z3}. Thus, this field has nine elements. A multiplication table for this field can be obtained from Table 13.1 by replacing i by x. (Why does this work?)

Unique Factorization in Z[x] As a further application of the ideas presented in this chapter, we next prove that Z[x] has an important factorization property. In Chapter 18, we will study this property in greater depth. The first proof of Theorem 17.6 was given by Gauss. In reading this theorem and its proof, keep in mind that the units in Z[x] are precisely f(x) 5 1 and f(x) 5 21 (see Exercise 25 in Chapter 12), the irreducible polynomials of degree 0 over Z are precisely those of the form f(x) 5 p and f(x) 5 2p where p is a prime, and every nonconstant polynomial from Z[x] that is irreducible over Z is primitive (see Exercise 3). Theorem 17.6 Unique Factorization in Z[x] Every polynomial in Z[x] that is not the zero polynomial or a unit in Z[x] can be written in the form b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x), where the bi’s are irreducible polynomials of degree 0, and the pi(x)’s are irreducible polynomials of positive degree. Furthermore, if b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ? qn(x), where the b’s and c’s are irreducible polynomials of degree 0, and the p(x)’s and q(x)’s are irreducible polynomials of positive degree, then s 5 t, m 5 n, and, after renumbering the c’s and q(x)’s, we have bi 5 6ci for i 5 1, . . . , s; and pi(x) 5 6qi(x) for i 5 1, . . . , m.

PROOF Let f(x) be a nonzero, nonunit polynomial from Z[x]. If deg f(x) 5 0, then f(x) is constant and the result follows from the Fundamental Theorem of Arithmetic. If deg f(x) . 0, let b denote the content of f(x), and let b1b2 ? ? ? bs be the factorization of b as a product of primes. Then, f(x) 5 b1b2 ? ? ? bs f1(x), where f1(x) belongs to Z[x], is

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primitive and deg f1(x) 5 deg f(x). Thus, to prove the existence portion of the theorem, it suffices to show that a primitive polynomial f(x) of positive degree can be written as a product of irreducible polynomials of positive degree. We proceed by induction on deg f(x). If deg f(x) 5 1, then f(x) is already irreducible and we are done. Now suppose that every primitive polynomial of degree less than deg f(x) can be written as a product of irreducibles of positive degree. If f(x) is irreducible, there is nothing to prove. Otherwise, f(x) 5 g(x)h(x), where both g(x) and h(x) are primitive and have degree less than that of f(x). Thus, by induction, both g(x) and h(x) can be written as a product of irreducibles of positive degree. Clearly, then, f(x) is also such a product. To prove the uniqueness portion of the theorem, suppose that f(x) 5 b1b2 ? ? ? bs p1(x)p2(x) ? ? ? pm(x) 5 c1c2 ? ? ? ct q1(x)q2(x) ? ? ? qn(x), where the b’s and c’s are irreducible polynomials of degree 0, and the p(x)’s and q(x)’s are irreducible polynomials of positive degree. Let b 5 b1b2 ? ? ? bs and c 5 c1c2 ? ? ? ct. Since the p(x)’s and q(x)’s are primitive, it follows from Gauss’s Lemma that p1(x)p2(x) ? ? ? pm(x) and q1(x)q2(x) ? ? ? qn(x) are primitive. Hence, both b and c must equal plus or minus the content of f(x) and, therefore, are equal in absolute value. It then follows from the Fundamental Theorem of Arithmetic that s 5 t and, after renumbering, bi 5 6ci for i 5 1, 2, . . . , s. Thus, by canceling the constant terms in the two factorizations for f(x), we have p1(x)p2(x) ? ? ? pm(x) 5 6q1(x) q2(x) ? ? ? qn(x). Now, viewing the p(x)’s and q(x)’s as elements of Q[x] and noting that p1(x) divides q1(x) ? ? ? qn(x), it follows from Corollary 2 of Theorem 17.5 and induction (see Exercise 27) that p1(x) | qi(x) for some i. By renumbering, we may assume i 5 1. Then, since q1(x) is irreducible, we have q1(x) 5 (r/s)p1(x), where r, s [ Z. However, because both q1(x) and p1(x) are primitive, we must have r/s 5 61. So, q1(x) 5 6p1(x). Also, after canceling, we have p2(x) ? ? ? pm(x) 5 6q2(x) ? ? ? qn(x). Now, we may repeat the argument above with p2(x) in place of p1(x). If m , n, after m such steps we would have 1 on the left and a nonconstant polynomial on the right. Clearly, this is impossible. On the other hand, if m . n, after n steps we would have 61 on the right and a nonconstant polynomial on the left— another impossibility. So, m 5 n and pi(x) 5 6qi(x) after suitable renumbering of the q(x)’s.

Weird Dice: An Application of Unique Factorization EXAMPLE 12 Consider an ordinary pair of dice whose faces are labeled 1 through 6. The probability of rolling a sum of 2 is 1/36, the probability of rolling a sum of 3 is 2/36, and so on. In a 1978 issue of

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Scientific American [1], Martin Gardner remarked that if one were to label the six faces of one cube with the integers 1, 2, 2, 3, 3, 4 and the six faces of another cube with the integers 1, 3, 4, 5, 6, 8, then the probability of obtaining any particular sum with these dice (called Sicherman dice) would be the same as the probability of rolling that sum with ordinary dice (that is, 1/36 for a 2, 2/36 for a 3, and so on). See Figure 17.1. In this example, we show how the Sicherman labels can be derived, and that they are the only possible such labels besides 1 through 6. To do so, we utilize the fact that Z[x] has the unique factorization property.

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Figure 17.1

To begin, let us ask ourselves how we may obtain a sum of 6, say, with an ordinary pair of dice. Well, there are five possibilities for the two faces: (5, 1), (4, 2), (3, 3), (2, 4), and (1, 5). Next we consider the product of the two polynomials created by using the ordinary dice labels as exponents: (x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x). Observe that we pick up the term x6 in this product in precisely the following ways: x5 ? x1, x4 ? x2 , x3 ? x3, x2 ? x4, x1 ? x5. Notice the correspondence between pairs of labels whose sums are 6 and pairs of terms whose products are x6. This correspondence is one-to-one, and it is valid for all sums and all dice—including the Sicherman dice and any other dice that yield the desired probabilities. So, let a1, a2, a3, a4, a5, a6 and b1, b2, b3, b4, b5, b6 be any two lists of positive integer labels for the faces of a pair of cubes with the property that the probability of rolling any particular sum with these dice (let us call them weird dice) is the same as the probability of rolling that sum with ordinary dice labeled 1 through 6. Using our observation about products of polynomials, this means that (x6 1 x5 1 x4 1 x3 1 x2 1 x)(x6 1 x5 1 x4 1 x3 1 x2 1 x) 5 (xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6) ? (xb1 1 xb2 1 xb3 1 xb4 1 xb5 1 xb6). (1)

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Now all we have to do is solve this equation for the a’s and b’s. Here is where unique factorization in Z[x] comes in. The polynomial x6 1 x5 1 x4 1 x3 1 x2 1 x factors uniquely into irreducibles as x(x 1 1)(x2 1 x 1 1)(x2 2 x 1 1) so that the left-hand side of Equation (1) has the irreducible factorization x2(x 1 1)2(x2 1 x 1 1)2(x2 2 x 1 1)2. So, by Theorem 17.6, this means that these factors are the only possible irreducible factors of P(x) 5 xa1 1 xa2 1 xa3 1 xa4 1 xa5 1 xa6. Thus, P(x) has the form xq(x 1 1)r(x2 1 x 1 1)t(x2 2 x 1 1)u, where 0 # q, r, t, u # 2. To restrict further the possibilities for these four parameters, we evaluate P(1) in two ways. P(1) 5 1a1 1 1a2 1 ? ? ? 1 1a6 5 6 and P(1) 5 1q2r3t1u. Clearly, this means that r 5 1 and t 5 1. What about q? Evaluating P(0) in two ways shows that q 2 0. On the other hand, if q 5 2, the smallest possible sum one could roll with the corresponding labels for dice would be 3. Since this violates our assumption, we have now reduced our list of possibilities for q, r, t, and u to q 5 1, r 5 1, t 5 1, and u 5 0, 1, 2. Let’s consider each of these possibilities in turn. When u 5 0, P(x) 5 x4 1 x3 1 x3 1 x2 1 x2 1 x, so the die labels are 4, 3, 3, 2, 2, 1—a Sicherman die. When u 5 1, P(x) 5 x6 1 x5 1 x4 1 x3 1 x2 1 x, so the die labels are 6, 5, 4, 3, 2, 1—an ordinary die. When u 5 2, P(x) 5 x8 1 x6 1 x5 1 x4 1 x3 1 x, so the die labels are 8, 6, 5, 4, 3, 1—the other Sicherman die. This proves that the Sicherman dice do give the same probabilities as ordinary dice and that they are the only other pair of dice that have this property.

Exercises No matter how good you are at something, there’s always about a million people better than you. HOMER SIMPSON

1. Verify the assertion made in Example 2. 2. Suppose that D is an integral domain and F is a field containing D. If f(x) [ D[x] and f(x) is irreducible over F but reducible over D, what can you say about the factorization of f(x) over D?

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3. Show that a nonconstant polynomial from Z[x] that is irreducible over Z is primitive. (This exercise is referred to in this chapter.) 4. Suppose that f(x) 5 xn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]. If r is rational and x 2 r divides f(x), show that r is an integer. 5. Let F be a field and let a be a nonzero element of F. a. If af(x) is irreducible over F, prove that f(x) is irreducible over F. b. If f(ax) is irreducible over F, prove that f(x) is irreducible over F. c. If f(x 1 a) is irreducible over F, prove that f(x) is irreducible over F. d. Use part c to prove that 8x3 2 6x 1 1 is irreducible over Q. (This exercise is referred to in this chapter.) 6. Suppose that f(x) [ Zp[x] and is irreducible over Zp, where p is a prime. If deg f(x) 5 n, prove that Zp[x]/ f(x) is a field with pn elements. 7. Construct a field of order 25. 8. Construct a field of order 27. 9. Show that x3 1 x2 1 x 1 1 is reducible over Q. Does this fact contradict the corollary to Theorem 17.4? 10. Determine which of the polynomials below is (are) irreducible over Q. a. x5 1 9x4 1 12x2 1 6 b. x4 1 x 1 1 c. x4 1 3x2 1 3 d. x5 1 5x2 1 1 e. (5/2)x5 1 (9/2)x4 1 15x3 1 (3/7)x2 1 6x 1 3/14 11. Show that x4 1 1 is irreducible over Q but reducible over R. (This exercise is referred to in this chapter.) 12. Show that x2 1 x 1 4 is irreducible over Z11. 13. Let f(x) 5 x3 1 6 [ Z7[x]. Write f(x) as a product of irreducible polynomials over Z7. 14. Let f(x) 5 x3 1 x2 1 x 1 1 [ Z2[x]. Write f(x) as a product of irreducible polynomials over Z2. 15. Let p be a prime. a. Show that the number of reducible polynomials over Zp of the form x2 1 ax 1 b is p( p 1 1)/2. b. Determine the number of reducible quadratic polynomials over Zp. 16. Let p be a prime. a. Determine the number of irreducible polynomials over Zp of the form x2 1 ax 1 b. b. Determine the number of irreducible quadratic polynomials over Zp.

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17. Show that for every prime p there exists a field of order p2. 18. Prove that, for every positive integer n, there are infinitely many polynomials of degree n in Z[x] that are irreducible over Q. 19. Show that the field given in Example 11 in this chapter is isomorphic to the field given in Example 9 in Chapter 13. 20. Let f(x) [ Zp[x]. Prove that if f(x) has no factor of the form x2 1 ax 1 b, then it has no quadratic factor over Zp. 21. Find all monic irreducible polynomials of degree 2 over Z3. 22. Given that p is not the zero of a nonzero polynomial with rational coefficients, prove that p 2 cannot be written in the form ap 1 b, where a and b are rational. 23. Find all the zeros and their multiplicities of x5 1 4x4 1 4x3 2 x2 2 4x 1 1 over Z5. 24. Find all zeros of f(x) 5 3x2 1 x 1 4 over Z7 by substitution. Find all zeros of f(x) by using the Quadratic Formula (2b 6 "b2 24ac) ? (2a)21 (all calculations are done in Z7). Do your answers agree? Should they? Find all zeros of g(x) 5 2x2 1 x 1 3 over Z5 by substitution. Try the Quadratic Formula on g(x). Do your answers agree? State necessary and sufficient conditions for the Quadratic Formula to yield the zeros of a quadratic from Zp[x], where p is a prime greater than 2. 25. (Rational Root Theorem) Let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x]

26.

27.

28. 29. 30.

and an 2 0. Prove that if r and s are relatively prime integers and f(r/s) 5 0, then r | a0 and s | an. Let F be a field and f(x) [ F[x]. Show that, as far as deciding upon the irreducibility of f(x) over F is concerned, we may assume that f(x) is monic. (This assumption is useful when one uses a computer to check for irreducibility.) Let F be a field and let p(x), a1(x), a2(x), . . . , ak(x) [ F[x], where p(x) is irreducible over F. If p(x) | a1(x)a2(x) ? ? ? ak(x), show that p(x) divides some ai(x). (This exercise is referred to in the proof of Theorem 17.6.) Explain how the Mod p Irreducibility Test (Theorem 17.3) can be used to test members of Q[x] for irreducibility. Show that x4 1 1 is reducible over Zp for every prime p. (This exercise is referred to in this chapter.) If p is a prime, prove that xp21 2 xp22 1 xp23 2 ? ? ? 2 x 1 1 is irreducible over Q.

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31. Let F be a field and let p(x) be irreducible over F. If E is a field that contains F and there is an element a in E such that p(a) 5 0, show that the mapping f: F[x] → E given by f(x) → f(a) is a ring homomorphism with kernel p(x). (This exercise is referred to in Chapter 20.) 32. Prove that the ideal x2 1 1 is prime in Z[x] but not maximal in Z[x]. 33. Let F be a field and let p(x) be irreducible over F. Show that {a 1  p(x) | a [ F} is a subfield of F[x]/p(x) isomorphic to F. (This exercise is referred to in Chapter 20.) 34. Suppose there is a real number r with the property that r 1 1/r is an odd integer. Prove that r is irrational. 35. In the game of Monopoly, would the probabilities of landing on various properties be different if the game were played with Sicherman dice instead of ordinary dice? Why? 36. Carry out the analysis given in Example 12 for a pair of tetrahedrons instead of a pair of cubes. (Define ordinary tetrahedral dice as the ones labeled 1 through 4.) 37. Suppose in Example 12 that we begin with n (n . 2) ordinary dice each labeled 1 through 6, instead of just two. Show that the only possible labels that produce the same probabilities as n ordinary dice are the labels 1 through 6 and the Sicherman labels. 38. Show that one two-sided die labeled with 1 and 4 and another 18sided die labeled with 1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8 yield the same probabilities as an ordinary pair of cubes labeled 1 through 6. Carry out an analysis similar to that given in Example 12 to derive these labels.

Computer Exercises The experiment serves two purposes, often independent one from the other: it allows the observation of new facts, hitherto either unsuspected, or not yet well defined; and it determines whether a working hypothesis fits the world of observable facts. RENÉ J. DUBOS

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software implements the Mod p Irreducibility Test. Use it to test the polynomials in the examples given in this chapter and the polynomials given in Exercise 10 for irreducibility.

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2. Use software such as Mathematica, Maple, or GAP to express xn 2 1 as a product of irreducible polynomials with integer coefficients for n 5 4, 8, 12, and 20. On the basis of these data, make a conjecture about the coefficients of the irreducible factors of xn 2 1. Test your conjecture for n 5 105. Does your conjecture hold up? n 3. Use software such as Mathematica, Maple, or GAP to express x p 2 x as a product of irreducibles over Zp for several choices of the prime p and n. On the basis of these data, make a conjecture relating the n degrees of the irreducible factors of x p 2 x and n.

Reference 1. Martin Gardner, “Mathematical Games,” Scientific American 238/2 (1978): 19–32.

Suggested Readings Duane Broline, “Renumbering the Faces of Dice,” Mathematics Magazine 52 (1979): 312–315. In this article, the author extends the analysis we carried out in Example 12 to dice in the shape of Platonic solids. J. A. Gallian and D. J. Rusin, “Cyclotomic Polynomials and Nonstandard Dice,” Discrete Mathematics 27 (1979): 245–259. Here Example 12 is generalized to the case of n dice each with m labels for all n and m greater than 1. Randall Swift and Brian Fowler, “Relabeling Dice,” Mathematics Magazine 72 (1999): 204–208. The authors use the method presented in this chapter to derive positive integer labels for a pair of dice that are not six-sided but give the same probabilities for the sum of the faces as a pair of cubes labeled 1 through 6.

Serge Lang Lang’s Algebra changed the way graduate algebra is taught . . . . It has affected all subsequent graduate-level algebra books. Citation for the Steele Prize

SERGE LANG was a prolific mathematician, inspiring teacher, and political activist. He was born near Paris on May 19, 1927. His family moved to Los Angeles when he was a teenager. Lang received a B.A. in physics from Caltech in 1946 and a Ph.D. in mathematics from Princeton in 1951 under Emil Artin (see the biography in Chapter 19). His first permanent position was at Columbia University in 1955, but in 1971 Lang resigned his position at Columbia as a protest against Columbia’s handling of Vietnam antiwar protesters. He joined Yale University in 1972 and remained there until his retirement. Lang made significant contributions to number theory, algebraic geometry, differential geometry, and analysis. He wrote more than 120 research articles and 60 books. His most famous and influential book was his graduate-level Algebra. Lang was a prize-

winning teacher known for his extraordinary devotion to students. Lang often got into heated discussions about mathematics, the arts, and politics. In one incident, he threatened to hit a fellow mathematician with a bronze bust for not conceding it was selfevident that the Beatles were greater musicians than Beethoven. Among Lang’s honors were the Steele Prize for Mathematical Exposition from the American Mathematcial Society, the Cole Prize in Algebra (see Chapter 25), and election to the National Academy of Sciences. Lang died on September 25, 2005, at the age of 78. For more information about Lang, visit: http://wikipedia.org/wiki/ Serge_Lang

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Divisibility in 18 Integral Domains Give me a fruitful error anytime, full of seeds, bursting with its own corrections. You can keep your sterile truth for yourself. VILFREDO PARETO

Irreducibles, Primes In the preceding two chapters, we focused on factoring polynomials over the integers or a field. Several of those results—unique factorization in Z[x] and the division algorithm for F[x], for instance—are natural counterparts to theorems about the integers. In this chapter and the next, we examine factoring in a more abstract setting. Definition Associates, Irreducibles, Primes

Elements a and b of an integral domain D are called associates if a 5 ub, where u is a unit of D. A nonzero element a of an integral domain D is called an irreducible if a is not a unit and, whenever b, c [ D with a 5 bc, then b or c is a unit. A nonzero element a of an integral domain D is called a prime if a is not a unit and a | bc implies a | b or a | c.

Roughly speaking, an irreducible is an element that can be factored only in a trivial way. Notice that an element a is a prime if and only if a is a prime ideal. Relating the definitions above to the integers may seem a bit confusing, since in Chapter 0 we defined a positive integer to be a prime if it satisfies our definition of an irreducible, and we proved that a prime integer satisfies the definition of a prime in an integral domain (Euclid’s Lemma). The source of the confusion is that in the case of the integers, the concepts of irreducibles and primes are equivalent, but in general, as we will soon see, they are not. The distinction between primes and irreducibles is best illustrated by integral domains of the form Z["d] 5 {a 1 b"d | a, b [ Z}, where d is not 1 and is not divisible by the square of a prime. (These rings are of fundamental importance in number theory.) To analyze these rings, we need a convenient method of determining their units, irreducibles, and 322

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primes. To do this, we define a function N, called the norm, from Z["d] into the nonnegative integers by N(a 1 b"d) 5 |a2 2 db2|. We leave it to the reader (Exercise 1) to verify the following four properties: N(x) 5 0 if and only if x 5 0; N(xy) 5 N(x)N(y) for all x and y; x is a unit if and only if N(x) 5 1; and, if N(x) is prime, then x is irreducible in Z["d]. EXAMPLE 1 We exhibit an irreducible in Z["23] that is not prime. Here, N(a 1 b"23) 5 a2 1 3b2. Consider 1 1 "23. Suppose that we can factor this as xy, where neither x nor y is a unit. Then N(xy) 5 N(x)N(y) 5 N(1 1 "23) 5 4, and it follows that N(x) 5 2. But there are no integers a and b that satisfy a2 1 3b2 5 2. Thus, x or y is a unit and 1 1 "23 is an irreducible. To verify that it is not prime, we observe that (1 1 "23)(1 2 "23) 5 4 5 2 ? 2, so that 1 1 "23 divides 2 ? 2. On the other hand, for integers a and b to exist so that 2 5 (1 1 "23)(a 1 b"23) 5 (a 2 3b) 1 (a 1 b)"23, we must have a 2 3b 5 2 and a 1 b 5 0, which is impossible. Showing that an element of a ring of the form Z["d] is irreducible is more difficult when d . 1. The next example illustrates one method of doing this. The example also shows that the converse of the fourth property above for the norm is not true. That is, it shows that x may be irreducible even if N(x) is not prime. EXAMPLE 2 The element 7 is irreducible in the ring Z["5]. To verify this assertion, suppose that 7 5 xy, where neither x nor y is a unit. Then 49 5 N(7) 5 N(x) N(y), and since x is not a unit, we cannot have N(x) 5 1. This leaves only the case N(x) 5 7. Let x 5 a 1 b"5. Then there are integers a and b satisfying |a2 2 5b2| 5 7. This means that a2 2 5b2 5 67. Viewing this equation modulo 7 and trying all possible cases for a and b reveals that the only solutions are a 5 0 5 b. But this means that both a and b are divisible by 7, and this implies that |a2 2 5b2| 5 7 is divisible by 49, which is false. Example 1 raises the question of whether or not there is an integral domain containing a prime that is not an irreducible. The answer: no. Theorem 18.1 Prime Implies Irreducible In an integral domain, every prime is an irreducible.

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PROOF Suppose that a is a prime in an integral domain and a 5 bc. We must show that b or c is a unit. By the definition of prime, we know that a | b or a | c. Say at 5 b. Then 1b 5 b 5 at 5 (bc)t 5 b(ct) and, by cancellation, 1 5 ct. Thus, c is a unit. Recall that a principal ideal domain is an integral domain in which every ideal has the form a. The next theorem reveals a circumstance in which primes and irreducibles are equivalent. Theorem 18.2 PID Implies Irreducible Equals Prime In a principal ideal domain, an element is an irreducible if and only if it is a prime.

PROOF Theorem 18.1 shows that primes are irreducibles. To prove the converse, let a be an irreducible element of a principal ideal domain D and suppose that a | bc. We must show that a | b or a | c. Consider the ideal I 5 {ax 1 by | x, y [ D} and let d 5 I. Since a [ I, we can write a 5 dr, and because a is irreducible, d is a unit or r is a unit. If d is a unit, then I 5 D and we may write 1 5 ax 1 by. Then c 5 acx 1 bcy, and since a divides both terms on the right, a also divides c. On the other hand, if r is a unit, then a 5 d 5 I, and, because b [ I, there is an element t in D such that at 5 b. Thus, a divides b. It is an easy consequence of the respective division algorithms for Z and F[x], where F is a field, that Z and F[x] are principal ideal domains (see Exercise 41 in Chapter 14 and Theorem 16.3). Our next example shows, however, that one of the most familiar rings is not a principal ideal domain. EXAMPLE 3 We show that Z[x] is not a principal ideal domain. Consider the ideal I 5 {f(x) [ Z[x] | f(0) is even}. We claim that I is not of the form h(x). If this were so, there would be f(x) and g(x) in Z[x] such that 2 5 h(x)f(x) and x 5 h(x)g(x), since both 2 and x belong to I. By the degree rule (Exercise 17 in Chapter 16), 0 5 deg 2 5 deg h(x) 1 deg f(x), so that h(x) is a constant polynomial. To determine which constant, we observe that 2 5 h(1)f(1). Thus, h(1) 5 61 or 62. Since 1 is not in I, we must have h(x) 5 62. But then x 5 62g(x), which is nonsense. We have previously proved that the integral domains Z and Z[x] have important factorization properties: Every integer greater than 1 can be

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uniquely factored as a product of irreducibles (that is, primes), and every nonzero, nonunit polynomial can be uniquely factored as a product of irreducible polynomials. It is natural to ask whether all integral domains have this property. The question of unique factorization in integral domains first arose with the efforts to solve a famous problem in number theory that goes by the name Fermat’s Last Theorem.

Historical Discussion of Fermat’s Last Theorem There are infinitely many nonzero integers x, y, z that satisfy the equation x2 1 y2 5 z2. But what about the equation x3 1 y3 5 z3 or, more generally, xn 1 yn 5 zn, where n is an integer greater than 2 and x, y, z are nonzero integers? Well, no one has ever found a single solution of this equation, and for more than three centuries many have tried to prove that there is none. The tremendous effort put forth by the likes of Euler, Legendre, Abel, Gauss, Dirichlet, Cauchy, Kummer, Kronecker, and Hilbert to prove that there are no solutions to this equation has greatly influenced the development of ring theory. About a thousand years ago, Arab mathematicians gave an incorrect proof that there were no solutions when n 5 3. The problem lay dormant until 1637, when the French mathematician Pierre de Fermat (1601–1665) wrote in the margin of a book, “. . . it is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or, generally, any power above the second into two powers of the same degree: I have discovered a truly marvelous demonstration [of this general theorem] which this margin is too narrow to contain.” Because Fermat gave no proof, many mathematicians tried to prove the result. The case where n 5 3 was done by Euler in 1770, although his proof was incomplete. The case where n 5 4 is elementary and was done by Fermat himself. The case where n 5 5 was done in 1825 by Dirichlet, who had just turned 20, and by Legendre, who was past 70. Since the validity of the case for a particular integer implies the validity for all multiples of that integer, the next case of interest was n 5 7. This case resisted the efforts of the best mathematicians until it was done by Gabriel Lamé in 1839. In 1847, Lamé stirred excitement by announcing that he had completely solved the problem. His approach was to factor the expression xp 1 yp, where p is an odd prime, into (x 1 y)(x 1 ay) ? ? ? (x 1 a p21y),

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where a is the complex number cos(2p/p) 1 i sin(2p/p). Thus, his factorization took place in the ring Z[a] 5 {a0 1 a1a 1 ? ? ? 1 ap21a p21 |ai [ Z}. But Lamé made the mistake of assuming that, in such a ring, factorization into the product of irreducibles is unique. In fact, three years earlier, Ernst Eduard Kummer had proved that this is not always the case. Undaunted by the failure of unique factorization, Kummer began developing a theory to “save” factorization by creating a new type of number. Within a few weeks of Lamé’s announcement, Kummer had shown that Fermat’s Last Theorem is true for all primes of a special type. This proved that the theorem was true for all exponents less than 100, prime or not, except for 37, 59, 67, and 74. Kummer’s work has led to the theory of ideals as we know it today. Over the centuries, many proposed proofs have not held up under scrutiny. The famous number theorist Edmund Landau received so many of these that he had a form printed with “On page ____, lines ____ to ____, you will find there is a mistake.” Martin Gardner, “Mathematical Games” columnist of Scientific American, had postcards printed to decline requests from readers asking him to examine their proofs. Recent discoveries tying Fermat’s Last Theorem closely to modern mathematical theories gave hope that these theories might eventually lead to a proof. In March 1988, newspapers and scientific publications worldwide carried news of a proof by Yoichi Miyaoka (see Figure 18.1). Within weeks, however, Miyaoka’s proof was shown to be invalid. In June 1993, excitement spread through the mathematics community with the announcement that Andrew Wiles of Princeton University had proved Fermat’s Last Theorem (see Figure 18.2). The Princeton mathematics department chairperson was quoted as saying, “When we heard it, people started walking on air.” But once again a proof did not hold up under scrutiny. This story does have a happy ending. The mathematical community has agreed on the validity of the revised proof given by Wiles and Richard Taylor in September of 1994. In view of the fact that so many eminent mathematicians were unable to prove Fermat’s Last Theorem, despite the availability of the vastly powerful theories, it seems highly improbable that Fermat had a correct proof. Most likely, he made the error that his successors made of assuming that the properties of integers, such as unique factorization, carry over to integral domains in general.

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Figure 18.1

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Figure 18.2

Andrew Wiles

Unique Factorization Domains We now have the necessary terminology to formalize the idea of unique factorization. Definition Unique Factorization Domain (UFD)

An integral domain D is a unique factorization domain if 1. every nonzero element of D that is not a unit can be written as a product of irreducibles of D, and 2. the factorization into irreducibles is unique up to associates and the order in which the factors appear.

Another way to formulate part 2 of this definition is the following: If p1 n 1 p 2 n 2 ? ? ? pr n r and q1 m 1 q 2m 2 ? ? ? q s m s are two factorizations of some element as a product of irreducibles, where no two of the pi’s are associates and no two of the q j’s are associates, then r 5 s, and each p i is an associate of one and only one q j and ni = mj . Of course, the Fundamental Theorem of Arithmetic tells us that the ring of integers is a unique factorization domain, and Theorem 17.6 says that Z[x] is a unique factorization domain. In fact, as we shall soon

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see, most of the integral domains we have encountered are unique factorization domains. Before proving our next theorem, we need the ascending chain condition for ideals. Lemma Ascending Chain Condition for a PID In a principal ideal domain, any strictly increasing chain of ideals I1 , I2 , ? ? ? must be finite in length.

PROOF Let I1 , I2 , ? ? ? be a chain of strictly increasing ideals in an integral domain D, and let I be the union of all the ideals in this chain. We leave it as an exercise (Exercise 3) to verify that I is an ideal of D. Then, since D is a principal ideal domain, there is an element a in D such that I 5 a. Because a [ I and I 5
PROOF Let D be a principal ideal domain and let a0 be any nonzero nonunit in D. We will show that a0 is a product of irreducibles (the product might consist of only one factor). We begin by showing that a0 has at least one irreducible factor. If a0 is irreducible, we are done. Thus, we may assume that a0 5 b1a1, where neither b1 nor a1 is a unit and a1 is nonzero. If a1 is not irreducible, then we can write a1 5 b2a2, where neither b2 nor a2 is a unit and a2 is nonzero. Continuing in this fashion, we obtain a sequence b1, b2, . . . of elements that are not units in D and a sequence a0, a1, a2, . . . of nonzero elements of D with an 5 bn11an11 for each n. Hence, a0 , a1 , ? ? ? is a strictly increasing chain of ideals (see Exercise 5), which, by the preceding lemma, must be finite, say, a0 , a1 , ? ? ? , ar. In particular, ar is an irreducible factor of a0. This argument shows that every nonzero nonunit in D has at least one irreducible factor. Now write a0 5 p1c1, where p1 is irreducible and c1 is not a unit. If c1 is not irreducible, then we can write c1 5 p2c2, where p2 is irreducible and c2 is not a unit. Continuing in this fashion, we obtain, as before, a strictly increasing sequence a0 , c1 , c2 , ? ? ? , which must end

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in a finite number of steps. Let us say that the sequence ends with cs. Then cs is irreducible and a0 5 p1p2 ? ? ? pscs, where each pi is also irreducible. This completes the proof that every nonzero nonunit of a principal ideal domain is a product of irreducibles. It remains to be shown that the factorization is unique up to associates and the order in which the factors appear. To do this, suppose that some element a of D can be written a 5 p1p2 ? ? ? pr 5 q1q2 ? ? ? qs, where the p’s and q’s are irreducible and repetition is permitted. We induct on r. If r 5 1, then a is irreducible and, clearly, s 5 1 and p1 5 q1. So we may assume that any element that can be expressed as a product of fewer than r irreducible factors can be so expressed in only one way (up to order and associates). Since p1 divides q1q2 ? ? ? qs, it must divide some qi (see Exercise 29), say, p1 | q1. Then, q1 5 up1, where u is a unit of D. Since up1p2 ? ? ? pr 5 uq1q2 ? ? ? qs 5 q1(uq2) ? ? ? qs and up1 5 q1, we have, by cancellation, p2 ? ? ? pr 5 (uq2) ? ? ? qs. The induction hypothesis now tells us that these two factorizations are identical up to associates and the order in which the factors appear. Hence, the same is true about the two factorizations of a. In the existence portion of the proof of Theorem 18.3, the only way we used the fact that the integral domain D is a principal ideal domain was to say that D has the property that there is no infinite, strictly increasing chain of ideals in D. An integral domain with this property is called a Noetherian domain, in honor of Emmy Noether, who inaugurated the use of chain conditions in algebra. Noetherian domains are of the utmost importance in algebraic geometry. One reason for this is that, for many important rings R, the polynomial ring R[x] is a Noetherian domain but not a principal ideal domain. One such example is Z[x]. In particular, Z[x] shows that a UFD need not be a PID (see Example 3). As an immediate corollary of Theorem 18.3, we have the following fact.

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Corollary F[x] Is a UFD Let F be a field. Then F[x] is a unique factorization domain.

PROOF By Theorem 16.3, F[x] is a principal ideal domain. So, F[x] is a unique factorization domain, as well. As an application of the preceding corollary, we give an elegant proof, due to Richard Singer, of Eisenstein’s Criterion (Theorem 17.4). EXAMPLE 4 Let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0 [ Z[x], and suppose that p is prime such that p B an, p | an21, . . . , p | a0

and

p2 B a0.

We will prove that f(x) is irreducible over Q. If f(x) is reducible over Q, we know by Theorem 17.2 that there exist elements g(x) and h(x) in Z[x] such that f(x) 5 g(x)h(x) and 1 # deg g(x) , n and 1 # deg h (x) , n. Let f (x), g(x), and h(x) be the polynomials in Zp[x] obtained from f (x), g(x), and h(x) by reducing all coefficients modulo p. Then, since p divides all the coefficients of f (x) except an, we have anxn 5 f (x) 5 g(x) h(x). Since Zp is a field, Zp[x] is a unique factorization domain. Thus, x | g(x) and x | h(x). So, g(0) 5 h(0) 5 0 and, therefore, p | g(0) and p | h(0). But, then, p2 | g(0)h(0) 5 f(0) 5 a0, which is a contradiction.

Euclidean Domains Another important kind of integral domain is a Euclidean domain. Definition Euclidean Domain

An integral domain D is called a Euclidean domain if there is a function d (called the measure) from the nonzero elements of D to the nonnegative integers such that 1. d(a) # d(ab) for all nonzero a, b in D; and 2. if a, b [ D, b 2 0, then there exist elements q and r in D such that a 5 bq 1 r, where r 5 0 or d(r) , d(b).

EXAMPLE 5 The ring Z is a Euclidean domain with d(a) 5 |a| (the absolute value of a).

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EXAMPLE 6 Let F be a field. Then F[x] is a Euclidean domain with d( f(x)) 5 deg f(x) (see Theorem 16.2). Examples 5 and 6 illustrate just one of many similarities between the rings Z and F[x]. Additional similarities are summarized in Table 18.1. Table 18.1 Similarities Between Z and F[x] Z

F[x]

Form of elements: an10n 1 an2110n21 1 ? ? ? 1 a110 1 a0 Euclidean domain: d(a) 5 |a| Units: a is a unit if and only if |a| 5 1 Division algorithm: For a, b [ Z, b 2 0, there exist q, r [ Z such that a 5 bq 1 r, 0 # r , |b|

↔

PID: Every nonzero ideal I 5 a, where a 2 0 and |a| is minimum Prime: No nontrivial factors UFD: Every element is a “unique” product of primes

↔

↔

↔

↔ ↔

Form of elements: anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 Euclidean domain: d( f(x)) 5 deg f(x) Units: f(x) is a unit if and only if deg f(x) 5 0 Division algorithm: For f(x), g(x) [ F[x], g(x) 2 0, there exist q(x), r(x) [ F[x] such that f(x) 5 g(x)q(x) 1 r(x), 0 # deg r(x) , deg g(x) or r(x) 5 0 PID: Every nonzero ideal I 5 f(x), where deg f(x) is minimum Irreducible: No nontrivial factors UFD: Every element is a “unique” product of irreducibles

EXAMPLE 7 The ring of Gaussian integers Z[i] 5 {a 1 bi | a, b [ Z} is a Euclidean domain with d(a 1 bi) 5 a2 1 b2. Unlike the previous two examples, in this example the function d does not obviously satisfy the necessary conditions. That d(x) # d(xy) for x, y [ Z[i] follows directly from the fact that d(xy) 5 d(x)d(y) (Exercise 7). To verify that condition 2 holds, observe that if x, y [ Z[i] and y 2 0, then xy21 [ Q[i], the field of quotients of Z[i] (Exercise 57 in Chapter 15). Say xy21 5 s 1 ti, where s, t [ Q. Now let m be the integer nearest s, and let n be the integer nearest t. (These integers may not be uniquely determined, but that does not matter.) Thus, |m 2 s| # 1/2 and |n 2 t| # 1/2. Then xy21 5 s 1 ti 5 (m 2 m 1 s) 1 (n 2 n 1 t)i 5 (m 1 ni) 1 [(s 2 m) 1 (t 2 n)i].

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So, x 5 (m 1 ni)y 1 [(s 2 m) 1 (t 2 n)i]y. We claim that the division condition of the definition of a Euclidean domain is satisfied with q 5 m 1 ni and r 5 [(s 2 m) 1 (t 2 n)i]y. Clearly, q belongs to Z[i], and since r 5 x 2 qy, so does r. Finally, d(r) 5 d([(s 2 m) 1 (t 2 n)i])d(y) 5 [(s 2 m)2 1 (t 2 n)2]d(y) #a

1 1 1 b d(y) , d(y). 4 4

Theorem 18.4 ED (Euclidean Domain) Implies PID Every Euclidean domain is a principal ideal domain.

PROOF Let D be a Euclidean domain and I a nonzero ideal of D. Among all the nonzero elements of I, let a be such that d(a) is a minimum. Then I 5 a. For, if b [ I, there are elements q and r such that b 5 aq 1 r, where r 5 0 or d(r) , d(a). But r 5 b 2 aq [ I, so d(r) cannot be less than d(a). Thus, r 5 0 and b [ a. Finally, the zero ideal is 0. Although it is not easy to verify, we remark that there are principal ideal domains that are not Euclidean domains. The first such example was given by T. Motzkin in 1949. A more accessible account of Motzkin’s result can be found in [2]. As an immediate consequence of Theorems 18.3 and 18.4, we have the following important result. Corollary ED Implies UFD Every Euclidean domain is a unique factorization domain.

We may summarize our theorems and remarks as follows: ED ⇒ PID ⇒ UFD UFD ⇒ / PID ⇒ / ED

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(You can remember these implications by listing the types alphabetically.) In Chapter 17, we proved that Z[x] is a unique factorization domain. Since Z is a unique factorization domain, the next theorem is a broad generalization of this fact. The proof is similar to that of the special case, and we therefore omit it. Theorem 18.5 D a UFD Implies D[x] a UFD If D is a unique factorization domain, then D[x] is a unique factorization domain.

We conclude this chapter with an example of an integral domain that is not a unique factorization domain. EXAMPLE 8 The ring Z["2 5] 5 {a 1 b"2 5 | a, b [ Z} is an integral domain but not a unique factorization domain. It is straightforward that Z["2 5] is an integral domain (see Exercise 9 in Chapter 13). To verify that unique factorization does not hold, we mimic the method used in Example 1 with N(a 1 b"2 5) 5 a2 1 5b2. Since N(xy) 5 N(x)N(y) and N(x) 5 1 if and only if x is a unit (see Exercise 1), it follows that the only units of Z["2 5] are 61. Now consider the following factorizations: 46 5 2 ? 23, 46 5 (1 1 3"2 5)(1 2 3"2 5).

We claim that each of these four factors is irreducible over Z["2 5]. Suppose that, say, 2 5 xy, where x, y [ Z["2 5] and neither is a unit. Then 4 5 N(2) 5 N(x)N(y) and, therefore, N(x) 5 N(y) 5 2, which is impossible. Likewise, if 23 5 xy were a nontrivial factorization, then N(x) 5 23. Thus, there would be integers a and b such that a2 1 5b2 5 23. Clearly, no such integers exist. The same argument applies to 1 6 3"2 5. In light of Examples 7 and 8, one can’t help but wonder for which d , 0 is Z["d] a unique factorization domain. The answer is only when d 5 21 or 22 (see [1, p. 297]). The case where d 5 21 was first proved, naturally enough, by Gauss.

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Exercises I tell them that if they will occupy themselves with the study of mathematics they will find in it the best remedy against lust of the flesh. THOMAS MANN, The Magic Mountain

1. For the ring Z["d] 5 {a 1 b"d | a, b [ Z}, where d 2 1 and d is not divisible by the square of a prime, prove that the norm N(a 1 b"d) 5 |a2 2 db2| satisfies the four assertions made preceding Example 1. (This exercise is referred to in this chapter.) 2. In an integral domain, show that a and b are associates if and only if a 5 b. 3. Show that the union of a chain I1 , I2 , ? ? ? of ideals of a ring R is an ideal of R. (This exercise is referred to in this chapter.) 4. In an integral domain, show that the product of an irreducible and a unit is an irreducible. 5. Suppose that a and b belong to an integral domain, b 2 0, and a is not a unit. Show that ab is a proper subset of b. (This exercise is referred to in this chapter.) 6. Let D be an integral domain. Define a , b if a and b are associates. Show that this defines an equivalence relation on D. 7. In the notation of Example 7, show that d(xy) 5 d(x)d(y). 8. Let D be a Euclidean domain with measure d. Prove that u is a unit in D if and only if d(u) 5 d(1). 9. Let D be a Euclidean domain with measure d. Show that if a and b are associates in D, then d(a) 5 d(b). 10. Let D be a principal ideal domain and let p [ D. Prove that p is a maximal ideal in D if and only if p is irreducible. 11. Trace through the argument given in Example 7 to find q and r in Z[i] such that 3 2 4i 5 (2 1 5i)q 1 r and d(r) , d(2 1 5i). 12. Let D be a principal ideal domain. Show that every proper ideal of D is contained in a maximal ideal of D. 13. In Z["2 5], show that 21 does not factor uniquely as a product of irreducibles. 14. Show that 1 2 i is an irreducible in Z[i]. 15. Show that Z["2 6] is not a unique factorization domain. (Hint: Factor 10 in two ways.) Why does this show that Z["2 6] is not a principal ideal domain? 16. Give an example of a unique factorization domain with a subdomain that does not have a unique factorization.

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17. In Z[i], show that 3 is irreducible but 2 and 5 are not. 18. Prove that 7 is irreducible in Z["6], even though N(7) is not prime. 19. Prove that if p is a prime in Z that can be written in the form a2 1 b2, then a 1 bi is irreducible in Z[i]. Find three primes that have this property and the corresponding irreducibles. 20. Prove that Z["2 3] is not a principal ideal domain. 21. In Z["2 5], prove that 1 1 3"2 5 is irreducible but not prime. 22. In Z["5], prove that both 2 and 1 1 "5 are irreducible but not prime. 23. Prove that Z["5] is not a unique factorization domain. 24. Let F be field. Show that in F[x] a prime ideal is a maximal ideal. 25. Let d be an integer less than 21 that is not divisible by the square of a prime. Prove that the only units of Z["d] are 11 and 21. 26. If a and b belong to Z["d], where d is not divisible by the square of a prime and ab is a unit, prove that a and b are units. 27. Prove or disprove that if D is a principal ideal domain, then D[x] is a principal ideal domain. 28. Determine the units in Z[i]. 29. Let p be a prime in an integral domain. If p | a1a2 ? ? ? an, prove that p divides some ai. (This exercise is referred to in this chapter.) 30. Show that 3x2 1 4x 1 3 [ Z5[x] factors as (3x 1 2)(x 1 4) and (4x 1 1)(2x 1 3). Explain why this does not contradict the corollary of Theorem 18.3. 31. Let D be a principal ideal domain and p an irreducible element of D. Prove that D/p is a field. 32. Show that an integral domain with the property that every strictly decreasing chain of ideals I1 . I2 . ? ? ? must be finite in length is a field. 33. An ideal A of a commutative ring R with unity is said to be finitely generated if there exist elements a1, a2, . . . , an of A such that A 5 a1, a2, . . . , an. An integral domain R is said to satisfy the ascending chain condition if every strictly increasing chain of ideals I1 , I2 , ? ? ? must be finite in length. Show that an integral domain R satisfies the ascending chain condition if and only if every ideal of R is finitely generated. 34. Prove or disprove that a subdomain of a Euclidean domain is a Euclidean domain. 35. Show that for any nontrivial ideal I of Z[i], Z[i]/I is finite. 36. Find the inverse of 1 1 "2 in Z["2]. What is the multiplicative order of 1 1 "2?

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37. In Z["2 7], show that N(6 1 2"2 7) 5 N(1 1 3"2 7) but 6 1 2"2 7 and 1 1 3"2 7 are not associates. 38. Let R 5 Z % Z % ? ? ? (the collection of all sequences of integers under componentwise addition and multiplication). Show that R has ideals I1, I2, I3, . . . with the property that I1 , I2 , I3 , ? ? ?. (Thus R does not have the ascending chain condition.) 39. Prove that in a unique factorization domain an element is irreducible if and only if it is prime. 40. Let F be a field and let R be the integral domain in F 3x4 generated by x 2 and x 3. (That is, R is contained in every integral domain in F 3x4 that contains x 2 and x 3.) Show that R is not a unique factorization domain. 41. Prove that for every field F, there are infinitely many irreducible elements in F 3x4 .

Computer Exercise I never use a computer. ANDREW WILES

Software for the computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. In the ring Z[i] (where i2 5 21), this software determines when a positive integer n is a prime in Z[i]. Run the program for several cases and formulate a conjecture based on your data.

Reference 1. H. M. Stark, An Introduction to Number Theory, Chicago, Ill.: Markham, 1970. 2. J. C. Wilson, “A Principal Ideal Ring That Is Not a Euclidean Ring,” Mathematics Magazine 46 (1973): 74–78.

Suggested Readings Oscar Campoli, “A Principal Ideal Domain That Is Not a Euclidean Domain,” The American Mathematical Monthly 95 (1988): 868–871. The author shows that {a 1 bu | a, b [ Z, u 5 (1 1 "219)/2} is a PID that is not an ED.

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Gina Kolata, “At Last, Shout of ‘Eureka!’ in Age-Old Math Mystery,” The New York Times, June 24, 1993. This front-page article reports on Andrew Wiles’s announced proof of Fermat’s Last Theorem. C. Krauthhammer, “The Joy of Math, or Fermat’s Revenge,” Time, April 18, 1988; 92. The demise of Miyaoka’s proof of Fermat’s Last Theorem is charmingly lamented. Sahib Singh, “Non-Euclidean Domains: An Example,” Mathematics Magazine 49 (1976): 243. This article gives a short proof that Z["2 n] 5 {a 1 b"2 n | a, b [ Z} is an integral domain that is not Euclidean when n . 2 and 2n mod 4 5 2 or 2n mod 4 5 3. Simon Singh and Kenneth Ribet, “Fermat’s Last Stand,” Scientific American 277 (1997): 68–73. This article gives an accessible description of Andrew Wiles’s proof of Fermat’s Last Theorem.

Suggested Video The Proof, Nova, http://shop.wgbh.org/product/show/7827 This documentary film shown on PBS’s NOVA program in 1997 chronicles the seven-year effort of Andrew Wiles to prove Fermat’s Last Theorem. It can be viewed in five segments at www.youtube.com.

Suggested Websites http://www.d.umn.edu/~jgallian This website has images of postage stamps featuring Fermat’s Last Theorem issued by France and the Czech Republic. http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem This website provides a concise history of the efforts to prove Fermat’s Last Theorem. It includes photographs, references and links.

Sophie Germain One of the very few women to overcome the prejudice and discrimination that tended to exclude women from the pursuit of higher mathematics in her time was Sophie Germain.

SOPHIE GERMAIN was born in Paris on April 1, 1776. She educated herself by reading the works of Newton and Euler in Latin and the lecture notes of Lagrange. In 1804, Germain wrote to Gauss about her work in number theory but used the pseudonym Monsieur LeBlanc because she feared that Gauss would not take seriously the efforts of a woman. Gauss gave Germain’s results high praise and a few years later, upon learning her true identity, wrote to her: But how to describe to you my admiration and astonishment at seeing my esteemed correspondent Mr. LeBlanc metamorphose himself into this illustrious personage who gives such a brilliant example of what I would find it difficult to believe. A taste for the abstract sciences in general and above all the mysteries of numbers is excessively rare: it is not a subject which strikes everyone; the enchanting charms of this sublime science reveal

themselves only to those who have the courage to go deeply into it. But when a person of the sex which, according to our customs and prejudices, must encounter infinitely more difficulties than men to familiarize herself with these thorny researches, succeeds nevertheless in surmounting these obstacles and penetrating the most obscure parts of them, then without doubt she must have the noblest courage, quite extraordinary talents, and a superior genius.

Germain is best known for her work on Fermat’s Last Theorem. She died on June 27, 1831, in Paris. For more information about Germain, visit: http://www-groups.dcs .st-and.ac.uk/~history

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Andrew Wiles For spectacular contributions to number theory and related fields, for major advances on fundamental conjectures, and for settling Fermat’s Last Theorem. Citation for the Wolf Prize

$5.00 each for public lectures on the proof. Scalpers received as much as $25.00 a ticket for the sold-out event. The bubble soon burst when experts Postage stamp issued by the had an opportunity to scrutinize Wiles’s Czech Republic in honor of manuscript. By December, Wiles released a Fermat’s Last Theorem. statement saying he was working to resolve a gap in the proof. In September of 1994, a paper by Wiles and Richard Taylor, a former IN 1993, ANDREW WILES of Princeton electristudent of Wiles, circumvented the gap in fied the mathematics community by announcthe original proof. Since then, many experts ing that he had proved Fermat’s Last Theorem have checked the proof and have found no after seven years of effort. His proof, which errors. One mathematician was quoted as ran 200 pages, relied heavily on ring theory saying, “The exuberance is back.” In 1997, and group theory. Because of Wiles’s solid Wiles’s proof was the subject of a PBS Nova reputation and because his approach was program. based on deep results that had already shed Wiles was born in 1953 in Cambridge, much light on the problem, many experts in England. He obtained his bachelor’s degree at the field believed that Wiles had succeeded Oxford and his doctoral degree at Cambridge where so many others had failed. Wiles’s University in 1980. He was a professor at Harachievement was reported in newspapers and vard before accepting his present position at magazines around the world. The New York Princeton in 1982, and he has received many Times ran a front-page story on it, and one TV prestigious awards. network announced it on the evening news. To find more information about Wiles, Wiles even made People magazine’s list of the visit: 25 most intriguing people of 1993! In San Francisco a group of mathematicians rented a http://www-groups.dcs 1200-seat movie theater and sold tickets for .st-and.ac.uk/~history/

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18 | Supplementary Exercises for Chapters 15–18

Supplementary Exercises for Chapters 15–18 The intelligence is proved not by ease of learning, but by understanding what we learn. JOSEPH WHITNEY

True/false questions for Chapters 15–18 are available on the Web at http://www.d.umn.edu/~jgallian/TF 1. Suppose that F is a field and there is a ring homomorphism from Z onto F. Show that F is isomorphic to Zp for some prime p. 2. Let Q["2] 5 {r 1 s"2 | r, s [ Q}. Determine all ring automorphisms of Q["2]. 3. (Second Isomorphism Theorem for Rings) Let A be a subring of R and let B be an ideal of R. Show that A > B is an ideal of A and that A/(A > B) is isomorphic to (A 1 B)/B. (Recall that A 1 B 5 {a 1 b | a [ A, b [ B}.) 4. (Third Isomorphism Theorem for Rings) Let A and B be ideals of a ring R with B # A. Show that A/B is an ideal of R/B and (R/B)/(A/B) is isomorphic to R/A. 5. Let f(x) and g(x) be irreducible polynomials over a field F. If f(x) and g(x) are not associates, prove that F[x]/ f(x)g(x) is isomorphic to F[x]/ f(x) % F[x]/g(x). 6. (Chinese Remainder Theorem for Rings) If R is a commutative ring and I and J are two proper ideals with I 1 J 5 R, prove that R/(I > J) is isomorphic to R/I % R/J. Explain why Exercise 5 is a special case of this theorem. 7. Prove that the set of all polynomials all of whose coefficients are even is a prime ideal in Z[x]. 8. Let R 5 Z["2 5] and let I 5 {a 1 b"2 5 | a, b [ Z, a 2 b is even}. Show that I is a maximal ideal of R. 9. Let R be a ring with unity and let a be a unit in R. Show that the mapping from R into itself given by x → axa21 is a ring automorphism.

10. Let a 1 b"2 5 belong to Z["2 5] with b 2 0. Show that 2 does not belong to a 1 b"2 5. 11. Show that Z[i]/2 1 i is a field. How many elements does it have? 12. Is the homomorphic image of a principal ideal domain a principal ideal domain? 13. In Z["2] 5 {a 1 b"2 | a, b [ Z}, show that every element of the form (3 1 2"2)n is a unit, where n is a positive integer.

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14. Let p be a prime. Show that there is exactly one ring homomorphism from Zm to Zpk if pk does not divide m, and exactly two ring homomorphisms from Zm to Zpk if pk does divide m. 15. Recall that a is an idempotent if a2 5 a. Show that if 1 1 k is an idempotent in Zn, then n 2 k is an idempotent in Zn. 16. Show that Zn (where n . 1) always has an even number of idempotents. (The number is 2d, where d is the number of distinct prime divisors of n.) 17. Show that the equation x2 1 y2 5 2003 has no solutions in the integers. 18. Prove that if both k and k 1 1 are idempotents in Zn and k 2 0, then n 5 2k. 19. Prove that x4 1 15x3 1 7 is irreducible over Q. 20. For any integers m and n, prove that the polynomial x3 1 (5m 1 1)x 1 5n 1 1 is irreducible over Z. 21. Prove that "2 is a maximal ideal in Z["2]. How many elements are in the ring Z["2]/"2? 22. Prove that Z["22] and Z["2] are unique factorization domains. (Hint: Mimic Example 7 in Chapter 18.) 23. Is 3 a maximal ideal in Z[i]? 24. Express both 13 and 5 1 i as products of irreducibles from Z[i]. 25. Let R 5 {a/b | a, b [ Z, 3 B b}. Prove that R is an integral domain. Find its field of quotients. 26. Give an example of a ring that contains a subring isomorphic to Z and a subring isomorphic to Z3. 27. Show that Z[i]/3 is not ring-isomorphic to Z3 % Z3. 28. For any n . 1, prove that R 5 e c 29. 30. 31. 32. 33.

a 0 d ` a, b P Z n f is ring0 b

isomorphic to Zn % Zn. Suppose that R is a commutative ring and I is an ideal of R. Prove that R[x]/I[x] is isomorphic to (R/I)[x]. Find an ideal I of Z8[x] such that the factor ring Z8[x]/I is a field. Find an ideal I of Z8[x] such that the factor ring Z8[x]/I is an integral domain but not a field. For any f(x) [ Zp[x], show that f(xp) 5 ( f(x)) p. Find an ideal I of Z[x] such that Z[x]/I is ring-isomorphic to Z3.

PA R T

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Fields

For online student resources, visit this textbook’s website at http://college.hmco.com/PIC/gallian7e

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19 Vector Spaces

Still round the corner there may wait A new road or a secret gate. J. R. R. TOLKIEN,

Lord of the Rings

Definition and Examples Abstract algebra has three basic components: groups, rings, and fields. Thus far we have covered groups and rings in some detail, and we have touched on the notion of a field. To explore fields more deeply, we need some rudiments of vector space theory that are covered in a linear algebra course. In this chapter, we provide a concise review of this material. Definition Vector Space

A set V is said to be a vector space over a field F if V is an Abelian group under addition (denoted by 1) and, if for each a [ F and v [ V, there is an element av in V such that the following conditions hold for all a, b in F and all u, v in V. 1. 2. 3. 4.

a(v 1 u) 5 av 1 au (a 1 b)v 5 av 1 bv a(bv) 5 (ab)v 1v 5 v

The members of a vector space are called vectors. The members of the field are called scalars. The operation that combines a scalar a and a vector v to form the vector av is called scalar multiplication. In general, we will denote vectors by letters from the end of the alphabet, such as u, v, w, and scalars by letters from the beginning of the alphabet, such as a, b, c.

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EXAMPLE 1 The set Rn 5 {(a1, a2, . . . , an) | ai [ R} is a vector space over R. Here the operations are the obvious ones. (a1, a2, . . . , an) 1 (b1, b2, . . . , bn) 5 (a1 1 b1, a2 1 b2, . . . , an 1 bn) and b(a1, a2, . . . , an) 5 (ba1, ba2, . . . , ban). EXAMPLE 2 The set M2(Q) of 2 3 2 matrices with entries from Q is a vector space over Q. The operations are c

a1 a2 b b a 1 b1 a2 1 b2 d 1 c 1 2d 5 c 1 d a3 a4 b3 b4 a3 1 b3 a4 1 b4

and bc

a1 a2 ba1 ba2 d. d 5 c ba3 ba4 a3 a4

EXAMPLE 3 The set Zp[x] of polynomials with coefficients from Zp is a vector space over Zp, where p is a prime. EXAMPLE 4 The set of complex numbers C 5 {a 1 bi | a, b [ R} is a vector space over R. The operations are the usual addition and multiplication of complex numbers. The next example is a generalization of Example 4. Although it appears rather trivial, it is of the utmost importance in the theory of fields. EXAMPLE 5 Let E be a field and let F be a subfield of E. Then E is a vector space over F. The operations are the operations of E.

Subspaces Of course, there is a natural analog of subgroup and subring. Definition Subspace

Let V be a vector space over a field F and let U be a subset of V. We say that U is a subspace of V if U is also a vector space over F under the operations of V.

EXAMPLE 6 The set {a2x2 1 a1x 1 a0 | a0, a1, a2 [ R} is a subspace of the vector space of all polynomials with real coefficients over R.

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EXAMPLE 7 Let V be a vector space over F and let v1, v2, . . . , vn be (not necessarily distinct) elements of V. Then the subset v1, v2, . . . , vn 5 {a1v1 1 a2v2 1 ? ? ? 1 anvn | a1, a2, . . . , an [ F} is called the subspace of V spanned by v1, v2, . . . , vn. Any summand of the form a1v1 1 a2v2 1 ? ? ? 1 anvn is called a linear combination of v1, v2, . . . , vn. If v1, v2, . . . , vn 5 V, we say that {v1, v2, . . . , vn} spans V.

Linear Independence The next definition is the heart of the theory. Definition Linearly Dependent, Linearly Independent

A set S of vectors is said to be linearly dependent over the field F if there are vectors v1, v2, . . . , vn from S and elements a1, a2, . . . , an from F, not all zero, such that a1v1 1 a2v2 1 ? ? ? 1 anvn 5 0. A set of vectors that is not linearly dependent over F is called linearly independent over F.

EXAMPLE 8 In R3 the vectors (1, 0, 0), (1, 0, 1), and (1, 1, 1) are linearly independent over R. To verify this, assume that there are real numbers a, b, and c such that a(1, 0, 0) 1 b(1, 0, 1) 1 c(1, 1, 1) 5 (0, 0, 0). Then (a 1 b 1 c, c, b 1 c) 5 (0, 0, 0). From this we see that a 5 b 5 c 5 0. Certain kinds of linearly independent sets play a crucial role in the theory of vector spaces. Definition Basis

Let V be a vector space over F. A subset B of V is called a basis for V if B is linearly independent over F and every element of V is a linear combination of elements of B.

The motivation for this definition is twofold. First, if B is a basis for a vector space V, then every member of V is a unique linear combination of the elements of B (see Exercise 19). Second, with every vector space spanned by finitely many vectors, we can use the notion of basis to associate a unique integer that tells us much about the vector space. (In fact, this integer and the field completely determine the vector space up to isomorphism—see Exercise 31.) a a1b d ` a, b P R f a1b b is a vector space over R (see Exercise 17). We claim that the set EXAMPLE 9 The set V 5 e c

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1 1 0 1 d, c d f is a basis for V over R. To prove that the set 1 0 1 1 B is linearly independent, suppose that there are real numbers a and b such that B5 ec

a c This gives c

1 1 0 1 0 0 d 1b c d 5 c d. 1 0 1 1 0 0

a a1b 0 d 5 c a1b b 0

0 d , so that a 5 b 5 0. On the other 0

hand, since every member of V has the form c

a a1b 1 1 0 1 d 5 ac d 1 bc d, a1b b 1 0 1 1

we see that B spans V. We now come to the main result of this chapter. Theorem 19.1 Invariance of Basis Size If {u1, u2, . . . , um} and {w1, w2, . . . , wn} are both bases of a vector space V over a field F, then m 5 n.

PROOF Suppose that m 2 n. To be specific, let us say that m , n. Consider the set {w1, u1, u2 , . . . , um}. Since the u’s span V, we know that w1 is a linear combination of the u’s, say, w1 5 a1u1 1 a2u2 1 ? ? ? 1 amum, where the a’s belong to F. Clearly, not all the a’s are 0. For convenience, say a1 2 0. Then {w1, u2, . . . , um} spans V (see Exercise 21). Next, consider the set {w1, w2, u2, . . . , um}. This time, w2 is a linear combination of w1, u2 , . . . , um, say, w2 5 b1w1 1 b2u2 1 . . . 1 bmum, where the b’s belong to F. Then at least one of b2, . . . , bm is nonzero, for otherwise the w’s are not linearly independent. Let us say b2 2 0. Then w1, w2, u3, . . . , um span V. Continuing in this fashion, we see that {w1, w2, . . . , wm} spans V. But then wm11 is a linear combination of w1, w2, . . . , wm and, therefore, the set {w1, . . . , wn} is not linearly independent. This contradiction finishes the proof. Theorem 19.1 shows that any two finite bases for a vector space have the same size. Of course, not all vector spaces have finite bases. However, there is no vector space that has a finite basis and an infinite basis (see Exercise 25).

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19 | Vector Spaces

Definition Dimension

A vector space that has a basis consisting of n elements is said to have dimension n. For completeness, the trivial vector space {0} is said to be spanned by the empty set and to have dimension 0.

Although it requires a bit of set theory that is beyond the scope of this text, it can be shown that every vector space has a basis. A vector space that has a finite basis is called finite dimensional; otherwise, it is called infinite dimensional.

Exercises Somebody who thinks logically is a nice contrast to the real world. THE LAW OF THUMB

1. Verify that each of the sets in Examples 1– 4 satisfies the axioms for a vector space. Find a basis for each of the vector spaces in Examples 1–4. 2. (Subspace Test) Prove that a nonempty subset U of a vector space V over a field F is a subspace of V if, for every u and u9 in U and every a in F, u 1 u9 [ U and au [ U. 3. Verify that the set in Example 6 is a subspace. Find a basis for this subspace. Is {x2 1 x 1 1, x 1 5, 3} a basis? 4. Verify that the set v1, v2, . . . , vn defined in Example 7 is a subspace. 5. Determine whether or not the set {(2, 21, 0), (1, 2, 5), (7, 21, 5)} is linearly independent over R. 6. Determine whether or not the set ec

2 1 0 1 1 1 d, c d, c df 1 0 1 2 1 1

is linearly independent over Z5. 7. If {u, v, w} is a linearly independent subset of a vector space, show that {u, u 1 v, u 1 v 1 w} is also linearly independent. 8. If S is a linearly dependent set of vectors, prove that one of the vectors in S is a linear combination of the others. 9. (Every finite spanning collection contains a basis.) If {v1, v2, . . . , vn} spans a vector space V, prove that some subset of the v’s is a basis for V. 10. (Every independent set is contained in a basis.) Let V be a finitedimensional vector space and let {v1, v2, . . . , vn} be a linearly

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independent subset of V. Show that there are vectors w1, w2, . . . , wm such that {v1, v2, . . . , vn, w1, . . . , wm} is a basis for V. 11. If V is a vector space over F of dimension 5 and U and W are subspaces of V of dimension 3, prove that U > W 2 {0}. Generalize. 12. Show that the solution set to a system of equations of the form a11x1 1 ? ? ? 1 a1nxn 5 0 a21x1 1 ? ? ? 1 a2nxn 5 0 ? ? ? ? ? ? ? ? ? am1x1 1 ? ? ? 1 amnxn 5 0, where the a’s are real, is a subspace of Rn. 13. Let V be the set of all polynomials over Q of degree 2 together with the zero polynomial. Is V a vector space over Q? 14. Let V 5 R3 and W 5 {(a, b, c) [ V | a2 1 b2 5 c2}. Is W a subspace of V? If so, what is its dimension? 15. Let V 5 R3 and W 5 {(a, b, c) [ V | a 1 b 5 c}. Is W a subspace of V? If so, what is its dimension? a b d ` a, b, c P Q f . Prove that V is a vector space b c over Q, and find a basis for V over Q. Verify that the set V in Example 9 is a vector space over R. Let P 5 {(a, b, c) | a, b, c [ R, a 5 2b 1 3c}. Prove that P is a subspace of R3. Find a basis for P. Give a geometric description of P. Let B be a subset of a vector space V. Show that B is a basis for V if and only if every member of V is a unique linear combination of the elements of B. (This exercise is referred to in this chapter and in Chapter 20.) If U is a proper subspace of a finite-dimensional vector space V, show that the dimension of U is less than the dimension of V. Referring to the proof of Theorem 19.1, prove that {w1, u2, . . . , um} spans V. If V is a vector space of dimension n over the field Zp, how many elements are in V? Let S 5 {(a, b, c, d) | a, b, c, d [ R, a 5 c, d 5 a 1 b}. Find a basis for S. Let U and W be subspaces of a vector space V. Show that U > W is a subspace of V and that U 1 W 5 {u 1 w | u [ U, w [ W} is a subspace of V.

16. Let V 5 e c 17. 18. 19.

20. 21. 22. 23. 24.

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25. If a vector space has one basis that contains infinitely many elements, prove that every basis contains infinitely many elements. (This exercise is referred to in this chapter.) 26. Let u 5 (2, 3, 1), v 5 (1, 3, 0), and w 5 (2, 23, 3). Since 12 u 2 2 1 3 v 2 6 w 5 (0, 0, 0), can we conclude that the set {u, v, w} is linearly dependent over Z7? 27. Define the vector space analog of group homomorphism and ring homomorphism. Such a mapping is called a linear transformation. Define the vector space analog of group isomorphism and ring isomorphism. 28. Let T be a linear transformation from V to W. Prove that the image of V under T is a subspace of W. 29. Let T be a linear transformation of a vector space V. Prove that {v [ V | T(v) 5 0}, the kernel of T, is a subspace of V. 30. Let T be a linear transformation of V onto W. If {v1, v2, . . . , vn} spans V, show that {T(v1), T(v2), . . . , T(vn)} spans W. 31. If V is a vector space over F of dimension n, prove that V is isomorphic as a vector space to Fn 5 {(a1, a2, . . . , an) | ai [ F}. (This exercise is referred to in this chapter.) 32. Let V be a vector space over an infinite field. Prove that V is not the union of finitely many proper subspaces of V.

Emil Artin For Artin, to be a mathematician meant to participate in a great common effort, to continue work begun thousands of years ago, to shed new light on old discoveries, to seek new ways to prepare the developments of the future. Whatever standards we use, he was a great mathematician. RICHARD BRAUER, Bulletin of the American Mathematical Society

EMIL ARTIN was one of the leading mathematicians of the 20th century and a major contributor to linear algebra and abstract algebra. Artin was born on March 3, 1898, in Vienna, Austria, and grew up in what was recently known as Czechoslovakia. He received a Ph.D. in 1921 from the University of Leipzig. From 1923 until he emigrated to America in 1937, he was a professor at the University of Hamburg. After one year at Notre Dame, Artin went to Indiana University. In 1946, he moved to Princeton, where he stayed until 1958. The last four years of his career were spent where it began, at Hamburg. Artin’s mathematics is both deep and broad. He made contributions to number theory, group theory, ring theory, field theory, Galois theory, geometric algebra, algebraic

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topology, and the theory of braids—a field he invented. Artin received the American Mathematical Society’s Cole Prize in number theory, and he solved one of the 23 famous problems posed by the eminent mathematician David Hilbert in 1900. Artin was an outstanding teacher of mathematics at all levels, from freshman calculus to seminars for colleagues. Many of his Ph.D. students as well as his son Michael have become leading mathematicians. Through his research, teaching, and books, Artin exerted great influence among his contemporaries. He died of a heart attack, at the age of 64, in 1962. For more information about Artin, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Olga Taussky-Todd “Olga Taussky-Todd was a distinguished and prolific mathematician who wrote about 300 papers.” EDITH LUCHINS AND MARY ANN MCLOUGHLIN,

Notices of the American Mathematical Society, 1996

OLGA TAUSSKY-TODD was born on August 30, 1906, in Olmütz in the Austro-Hungarian Empire. Taussky-Todd received her doctoral degree in 1930 from the University of Vienna. In the early 1930s she was hired as an assistant at the University of Göttingen to edit books on the work of David Hilbert. She also edited lecture notes of Emil Artin and assisted Richard Courant. She spent 1934 and 1935 at Bryn Mawr and the next two years at Girton College in Cambridge, England. In 1937, she taught at the University of London. In 1947, she moved to the United States and took a job at the National Bureau of Standards’ National Applied Mathematics Laboratory. In 1957, she became the first woman to teach at the California Institute of Technology as well as the first woman to receive tenure and a full professorship in mathematics, physics, or astronomy there. Thirteen Caltech Ph.D. students wrote their Ph.D. theses under her direction.

In addition to her influential contributions to linear algebra, Taussky-Todd did important work in number theory. Taussky-Todd received many honors and awards. She was elected a Fellow of the American Association for the Advancement of Science and vice president of the American Mathematical Society. In 1990, Caltech established an instructorship named in her honor. Taussky-Todd died on October 7, 1995, at the age of 89. For more information about TausskyTodd, visit: http://www-groups.dcs .st-and.ac.uk/~history http://www-scottlan .edu/lriddle/women/women.html

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20 Extension Fields

In many respects this [Kronecker’s Theorem] is the fundamental theorem of algebra. RICHARD A. DEAN,

Elements of Abstract Algebra

The Fundamental Theorem of Field Theory In our work on rings, we came across a number of fields, both finite and infinite. Indeed, we saw that Z3[x]/x2 1 1 is a field of order 9, whereas R[x]/x2 1 1 is a field isomorphic to the complex numbers. In the next three chapters, we take up, in a systematic way, the subject of fields. Definition Extension Field

A field E is an extension field of a field F if F # E and the operations of F are those of E restricted to F.

Cauchy’s observation in 1847 that R[x]/x2 1 1 is a field that contains a zero of x2 1 1 prepared the way for the following sweeping generalization of that fact. Theorem 20.1 Fundamental Theorem of Field Theory (Kronecker’s Theorem, 1887) Let F be a field and let f(x) be a nonconstant polynomial in F[x]. Then there is an extension field E of F in which f(x) has a zero.

PROOF Since F[x] is a unique factorization domain, f (x) has an irreducible factor, say, p(x). Clearly, it suffices to construct an extension field E of F in which p(x) has a zero. Our candidate for E is F[x]/ p(x). We already know that this is a field from Corollary 1 of Theorem 17.5. Also, since the mapping of f: F S E given by f(a) 5 a 1  p(x) is one-to-one and preserves both operations, E has a subfield isomorphic 354

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to F. We may think of E as containing F if we simply identify the coset a 1 p(x) with its unique coset representative a that belongs to F [that is, think of a 1  p(x) as just a and vice versa; see Exercise 33 in Chapter 17]. Finally, to show that p(x) has a zero in E, write p(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a0. Then, in E, x 1 p(x) is a zero of p(x), because p(x 1  p(x)) 5 an(x 1  p(x))n 1 an21(x 1  p(x))n21 1 ? ? ? 1 a0 5 an(xn 1  p(x)) 1 an21(xn21 1  p(x)) 1 ? ? ? 1 a0 5 anxn 1 an21xn21 1 ? ? ? 1 a0 1  p(x) 5 p(x) 1  p(x) 5 0 1  p(x). EXAMPLE 1 Let f(x) 5 x2 1 1 [ Q[x]. Then, viewing f(x) as an element of E[x] 5 (Q[x]/x2 1 1)[x], we have f(x 1 x2 1 1) 5 (x 1 x2 1 1)2 1 1 5 x2 1 x2 1 1 1 1 5 x2 1 1 1 x2 1 1 5 0 1 x2 1 1. Of course, the polynomial x2 1 1 has the complex number "21 as a zero, but the point we wish to emphasize here is that we have constructed a field that contains the rational numbers and a zero for the polynomial x2 1 1 by using only the rational numbers. No knowledge of complex numbers is necessary. Our method utilizes only the field we are given. EXAMPLE 2 Let f(x) 5 x5 1 2x2 1 2x 1 2 [ Z3[x]. Then, the irreducible factorization of f(x) over Z3 is (x2 1 1)(x3 1 2x 1 2). So, to find an extension E of Z3 in which f(x) has a zero, we may take E 5 Z3[x]/ x2 1 1, a field with nine elements, or E 5 Z3[x]/x3 1 2x 1 2, a field with 27 elements. Since every integral domain is contained in its field of quotients (Theorem 15.6), we see that every nonconstant polynomial with coefficients from an integral domain always has a zero in some field containing the ring of coefficients. The next example shows that this is not true for commutative rings in general. EXAMPLE 3 Let f(x) 5 2x 1 1 [ Z4[x]. Then f(x) has no zero in any ring containing Z4 as a subring, because if b were a zero in such a ring,

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then 0 5 2b 1 1, and therefore 0 5 2(2b 1 1) 5 2(2b) 1 2 5 (2 ? 2)b 1 2 5 0 ? b 1 2 5 2. But 0 2 2 in Z4.

Splitting Fields To motivate the next definition and theorem, let’s return to Example 1 for a moment. For notational convenience, in Q[x]/x2 1 1, let a 5 x 1 x2 1 1. Then, since a and 2a are both zeros of x2 1 1 in (Q[x]/ x2 1 1)[x], it should be the case that x2 1 1 5 (x 2 a)(x 1 a). Let’s check this out. First note that (x 2 a)(x 1 a) 5 x2 2 a2 5 x2 2 (x2 1 x2 1 1). At the same time, x2 1 x2 1 1 5 21 1 x2 1 1 and we have agreed to identify 21 and 21 1 x2 1 1, so (x 2 a)(x 1 a) 5 x2 2 (21) 5 x2 1 1. This shows that x2 1 1 can be written as a product of linear factors in some extension of Q. That was easy and you might argue coincidental. The polynomial given in Example 2 presents a greater challenge. Is there an extension of Z3 in which that polynomial factors as a product of linear factors? Yes, there is. But first a definition. Definition Splitting Field

Let E be an extension field of F and let f(x) [ F[x]. We say that f(x) splits in E if f(x) can be factored as a product of linear factors in E[x]. We call E a splitting field for f(x) over F if f(x) splits in E but in no proper subfield of E.

Note that a splitting field of a polynomial over a field depends not only on the polynomial but on the field as well. Indeed, a splitting field of f(x) over F is just a smallest extension field of F in which f(x) splits. The next example illustrates how a splitting field of a polynomial f (x) over field F depends on F. EXAMPLE 4 Consider the polynomial f(x) 5 x2 1 1 [ Q[x]. Since

x2 1 1 5 (x 1 "21)(x 2 "21), we see that f(x) splits in C, but a splitting

field over Q is Q(i) 5 {r 1 si | r, s [ Q}. A splitting field for x2 1 1 over R is C. Likewise, x2 2 2 [ Q[x] splits in R, but a splitting field over Q is Q("2) 5 {r 1 s "2 | r, s [ Q}.

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There is a useful analogy between the definition of a splitting field and the definition of an irreducible polynomial. Just as it makes no sense to say “f(x) is irreducible,” it makes no sense to say “E is a splitting field for f(x).” In each case, the underlying field must be specified; that is, one must say “f(x) is irreducible over F” and “E is a splitting field for f(x) over F.” The following notation is convenient. Let F be a field and let a1, a2, . . . , an be elements of some extension E of F. We use F(a1, a2, . . . , an) to denote the smallest subfield of E that contains F and the set {a1, a2, . . . , an}. It is an easy exercise to show that F(a1, a2, . . . , an) is the intersection of all subfields of E that contain F and the set {a1, a2, . . . , an}. Notice that if f(x) [ F[x] and f(x) factors as b(x 2 a1)(x 2 a2) ? ? ? (x 2 an) over some extension E of F, then F(a1, . . . , an) is a splitting field for f(x) over F in E. This notation appears to be inconsistent with the notation that we used in earlier chapters. For example, we denoted the set {a 1 b"2 | a, b [ Z} by Z["2] and the set {a 1 b"2 | a, b [ Q} by Q("2 ). The difference is that Z["2] is merely a ring, whereas Q("2) is a field. In general, parentheses are used when one wishes to indicate that the set is a field, although no harm would be done by using, say, Q["2] to denote {a 1 b"2 | a, b [ Q} if we were concerned with its ring properties only. Using parentheses rather than brackets simply conveys a bit more information about the set. Theorem 20.2 Existence of Splitting Fields Let F be a field and let f(x) be a nonconstant element of F[x]. Then there exists a splitting field E for f(x) over F.

PROOF We proceed by induction on deg f (x). If deg f (x) 5 1, then f (x) is linear. Now suppose that the statement is true for all fields and all polynomials of degree less than that of f (x). By Theorem 20.1, there is an extension E of F in which f(x) has a zero, say, a1. Then we may write f (x) 5 (x 2 a1)g(x), where g(x) [ E[x]. Since deg g(x) , deg f (x), by induction, there is a field K that contains E and all the zeros of g(x), say, a2, . . . , an. Clearly, then, a splitting field for f (x) over F is F(a1, a2, . . . , an). EXAMPLE 5 Consider f(x) 5 x4 2 x2 2 2 5 (x2 2 2)(x2 1 1)

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over Q. Obviously, the zeros of f (x) in C are 6"2 and 6i. So a splitting field for f(x) over Q is Q("2, i) 5 Q("2)(i) 5 {a 1 bi | a, b [ Q("2)}

5 {(a 1 b "2 ) 1 (c 1 d "2 )i | a, b, c, d [ Q}.

EXAMPLE 6 Consider f(x) 5 x2 1 x 1 2 over Z3. Then Z3(i) 5 {a 1 bi | a, b [ Z3} (see Example 9 in Chapter 13) is a splitting field for f(x) over Z3 because f(x) 5 [x 2 (1 1 i)][x 2 (1 2 i)]. At the same time, we know by the proof of Kronecker’s Theorem that the element x 1 x2 1 x 1 2 of F 5 Z3[x]/x2 1 x 1 2 is a zero of f(x). Since f(x) has degree 2, it follows from the Factor Theorem (Corollary 2 of Theorem 16.2) that the other zero of f(x) must also be in F. Thus, f(x) splits in F, and because F is a two-dimensional vector space over Z3 we know that F is also a splitting field of f(x) over Z3. But how do we factor f(x) in F? Factoring f(x) in F is confusing because we are using the symbol x in two distinct ways: It is used as a placeholder to write the polynomial f(x), and it is used to create the coset representatives of the elements of F. This confusion can be avoided by simply identifying the coset 1 1 x2 1 x 1 2 with the element 1 in Z3 and denoting the coset x 1 x2 1 x 1 2 by b. With this identification, the field Z3[x]/x2 1 x 1 2 can be represented as {0, 1, 2, b, 2b, b 1 1, 2b 1 1, b 1 2, 2b 1 2}. These elements are added and multiplied just as polynomials are, except that we use the observation that x2 1 x 1 2 1 x2 1 x 1 2 5 0 implies that b2 1 b 1 2 5 0, so that b2 5 2b 2 2 5 2b 1 1. For example, (2b 1 1)(b 1 2) 5 2b2 1 5b 1 2 5 2(2b 1 1) 1 5b 1 2 5 9b 1 4 5 1. To obtain the factorization of f(x) in F, we simply long divide, as follows: x 1 (b 1 1) x 2 b qx 2 1 x 1 2 x 2 2 bx (b 1 1)x 1 2 (b 1 1)x 2 (b 1 1)b (b 1 1)b 1 2 5 b2 1 b 1 2 5 0. So, x2 1 x 1 2 5 (x 2 b)(x 1 b 1 1). Thus, we have found two splitting fields for x2 1 x 1 2 over Z3, one of the form F(a) and one of the form F[x]/p(x) [where F 5 Z3 and p(x) 5 x2 1 x 1 2].

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The next theorem shows how the fields F(a) and F[x]/ p(x) are related in the case where p(x) is irreducible over F and a is a zero of p(x) in some extension of F. Theorem 20.3 F(a) < F[x]/p(x) Let F be a field and let p(x) [ F[x] be irreducible over F. If a is a zero of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/p(x). Furthermore, if deg p(x) 5 n, then every member of F(a) can be uniquely expressed in the form cn21an21 1 cn22an22 1 ? ? ? 1 c1a 1 c0, where c0 , c1, . . . , cn21 [ F.

PROOF Consider the function f from F[x] to F(a) given by f(f(x)) 5 f(a). Clearly, f is a ring homomorphism. We claim that Ker f 5 p(x). (This is Exercise 31 in Chapter 17.) Since p(a) 5 0, we have p(x) # Ker f. On the other hand, we know by Theorem 17.5 that p(x) is a maximal ideal in F[x]. So, because Ker f 2 F[x] [it does not contain the constant polynomial f(x) 5 1], we have Ker f 5 p(x). At this point it follows from the First Isomorphism Theorem for Rings and Corollary 1 of Theorem 17.5 that f(F[x]) is a subfield of F(a). Noting that f(F[x]) contains both F and a and recalling that F(a) is the smallest such field, we have F[x]/p(x) < f(F[x]) 5 F(a). The final assertion of the theorem follows from the fact that every element of F[x]/p(x) can be expressed uniquely in the form cn21x n21 1 ? ? ? 1 c0 1 p(x), where c0, . . . , cn21 [ F (see Exercise 23 in Chapter 16) and the natural isomorphism from F[x]/ p(x) to F(a) carries ckxk 1  p(x) to ckak. As an immediate corollary of Theorem 20.3, we have the following attractive result. Corollary F(a) < F(b) Let F be a field and let p(x) [ F[x] be irreducible over F. If a is a zero of p(x) in some extension E of F and b is a zero of p(x) in some extension E9 of F, then the fields F(a) and F(b) are isomorphic.

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PROOF From Theorem 20.3, we have F(a) < F[x]/p(x) < F(b). Recall that a basis for an n-dimensional vector space over a field F is a set of n vectors v1, v2, . . . , vn with the property that every member of the vector space can be expressed uniquely in the form a1v1 1 a2v2 1 ? ? ? 1 anvn, where the a’s belong to F (Exercise 19 in Chapter 19). So, in the language of vector spaces, the latter portion of Theorem 20.3 says that if a is a zero of an irreducible polynomial over F of degree n, then the set {1, a, . . . , an21} is a basis for F(a) over F. Theorem 20.3 often provides a convenient way of describing the elements of a field. EXAMPLE 7 Consider the irreducible polynomial f(x) 5 x6 2 2 6 over Q. Since " 2 is a zero of f(x), we know from Theorem 20.3 that the 6 set {1, 21/6, 22/6, 23/6, 24/6, 25/6} is a basis for Q(" 2) over Q. Thus, Q(" 2) 5 {a0 1 a121/6 1 a222/6 1 a323/6 1 a424/6 1 a525/6 | ai [ Q}. 6

This field is isomorphic to Q[x]/x6 2 2. In 1882, Ferdinand Lindemann (1852–1939) proved that p is not the zero of any polynomial in Q[x]. Because of this important result, Theorem 20.3 does not apply to Q(p) (see Exercise 11). In Example 6, we produced two splitting fields for the polynomial x2 1 x 1 2 over Z3. Likewise, it is an easy exercise to show that both Q[x]/x2 1 1 and Q(i) 5 {r 1 si | r, s [ Q} are splitting fields of the polynomial x2 1 1 over Q. But are these different-looking splitting fields algebraically different? Not really. We conclude our discussion of splitting fields by proving that splitting fields are unique up to isomorphism. To make it easier to apply induction, we will prove a more general result. We begin by observing first that any ring isomorphism f from F to F9 has a natural extension from F[x] to F9[x] given by cnx n 1 cn21xn21 1 ? ? ? 1 c1x 1 c0 S f(cn)x n 1 f(cn21)x n21 1 ? ? ? 1 f(c1)x 1 f(c0). Since this mapping agrees with f on F, it is convenient and natural to use f to denote this mapping as well. Lemma Let F be a field, let p(x) [ F[x] be irreducible over F, and let a be a zero of p(x) in some extension of F. If f is a field isomorphism from F to F9 and b is a zero of f(p(x)) in some extension of F9, then there is an isomorphism from F(a) to F9(b) that agrees with f on F and carries a to b.

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PROOF First observe that since p(x) is irreducible over F, f(p(x)) is irreducible over F9. It is straightforward to check that the mapping from F[x]/p(x) to F9[x]/f(p(x)) given by f(x) 1 p(x) S f(f(x)) 1 f(p(x)) is a field isomorphism. By a slight abuse of notation, we denote this mapping by f also. (If you object, put a bar over the f.) From the proof of Theorem 20.3, we know that there is an isomorphism a from F(a) to F[x]/p(x) that is the identity on F and carries a to x 1 p(x). Similarly, there is an isomorphism b from F9[x]/f(p(x)) to F9(b) that is the identity on F9 and carries x 1 f(p(x)) to b. Thus, bfa is the desired mapping from F(a) to F9(b) . See Figure 20.1. F'[x]/7 φ (p(x))8

F[x]/7 p(x)8

F(a)

φ

α

F

φ

F'(b) β

F'

Figure 20.1

Theorem 20.4 Extending f: F → F9 Let f be an isomorphism from a field F to a field F9 and let f(x) [ F[x]. If E is a splitting field for f(x) over F and E9 is a splitting field for f(f (x)) over F9, then there is an isomorphism from E to E9 that agrees with f on F.

PROOF We induct on deg f (x). If deg f (x) 5 1, then E 5 F and E9 5 F9, so that f itself is the desired mapping. If deg f (x) . 1, let p(x) be an irreducible factor of f (x), let a be a zero of p(x) in E, and let b be a zero of f(p(x)) in E9. By the preceding lemma, there is an isomorphism a from F(a) to F9(b) that agrees with f on F and carries a to b. Now write f (x) 5 (x 2 a)g(x), where g(x) [ F(a)[x]. Then E is a splitting field for g(x) over F(a) and E9 is a splitting field for a(g(x)) over F9(b). Since deg g(x) , deg f (x), there is an isomorphism from E to E9 that agrees with a on F(a) and therefore with f on F.

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Corollary Splitting Fields Are Unique Let F be a field and let f(x) [ F[x]. Then any two splitting fields of f(x) over F are isomorphic.

PROOF Suppose that E and E9 are splitting fields of f (x) over F. The result follows immediately from Theorem 20.4 by letting f be the identity from F to F. In light of the corollary above, we may refer to “the” splitting field of a polynomial over F without ambiguity. 6 6 Even though x6 2 2 has a zero in Q(" 2), it does not split in Q(" 2). The splitting field is easy to obtain, however. EXAMPLE 8 The Splitting Field of xn 2 a over Q Let a be a positive rational number and let v be a primitive nth root of unity (see Example 2 in Chapter 16). Then each of a1/n, va1/n, v2a1/n, . . . , vn21a1/n is a zero of xn 2 a in Q( "a , v). n

Zeros of an Irreducible Polynomial Now that we know that every nonconstant polynomial over a field splits in some extension, we ask whether irreducible polynomials must split in some special way. Yes, they do. To discover how, we borrow something whose origins are in calculus. Definition

Let f(x) 5 anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 belong to F[x]. The derivative of f(x), denoted by f9(x), is the polynomial nanx n21 1 (n 2 1)an21x n22 1 ? ? ? 1 a1 in F[x].

Notice that our definition does not involve the notion of a limit. The standard rules for handling sums and products of functions in calculus carry over to arbitrary fields as well. Lemma Properties of the Derivative Let f(x) and g(x) [ F[x] and let a [ F. Then 1. (f(x) 1 g(x))9 5 f 9(x) 1 g9(x) 2. (af(x))9 5 af 9(x) 3. (f(x)g(x))9 5 f(x)g9(x) 1 g(x)f 9(x).

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PROOF Parts 1 and 2 follow from straightforward applications of the definition. Using part 1 and induction on deg f (x), part 3 reduces to the special case in which f (x) 5 anx n. This also follows directly from the definition. Before addressing the question of the nature of the zeros of an irreducible polynomial, we establish a general result concerning zeros of multiplicity greater than 1. Such zeros are called multiple zeros. Theorem 20.5 Criterion for Multiple Zeros A polynomial f(x) over a field F has a multiple zero in some extension E if and only if f(x) and f9(x) have a common factor of positive degree in F[x].

PROOF If a is a multiple zero of f (x) in some extension E, then there is a g(x) in E[x] such that f(x) 5 (x 2 a)2g(x). Since f 9(x) 5 (x 2 a)2g9(x) 1 2(x 2 a)g(x), we see that f 9(a) 5 0. Thus x 2 a is a factor of both f(x) and f 9(x) in the extension E of F. Now if f(x) and f 9(x) have no common divisor of positive degree in F[x], there are polynomials h(x) and k(x) in F[x] such that f(x)h(x) 1 f 9(x)k(x) 5 1 (see Exercise 41 in Chapter 16). Viewing f(x)h(x) 1 f 9(x)k(x) as an element of E[x], we see also that x 2 a is a factor of 1. Since this is nonsense, f(x) and f 9(x) must have a common divisor of positive degree in F[x]. Conversely, suppose that f(x) and f 9(x) have a common factor of positive degree. Let a be a zero of the common factor. Then a is a zero of f(x) and f 9(x). Since a is a zero of f (x), there is a polynomial q(x) such that f (x) 5 (x 2 a)q(x). Then f 9(x) 5 (x 2 a)q9(x) 1 q(x) and 0 5 f 9(a) 5 q(a). Thus, x 2 a is a factor of q(x) and a is a multiple zero of f(x). Theorem 20.6 Zeros of an Irreducible Let f(x) be an irreducible polynomial over a field F. If F has characteristic 0, then f(x) has no multiple zeros. If F has characteristic p 2 0, then f(x) has a multiple zero only if it is of the form f(x) 5 g(x p) for some g(x) in F[x].

PROOF If f (x) has a multiple zero, then, by Theorem 20.5, f (x) and f 9(x) have a common divisor of positive degree in F[x]. Since the only divisor of positive degree of f(x) in F[x] is f(x) itself (up to associates), we see that f(x) divides f 9(x). Because a polynomial over a field cannot divide a polynomial of smaller degree, we must have f 9(x) 5 0.

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Now what does it mean to say that f 9(x) 5 0? If we write f(x) 5 anxn 1 an21x n21 1 ? ? ? 1 a1x 1 a0, then f9(x) 5 nanx n21 1 (n 2 1)an21x n22 1 ? ? ? 1 a1. Thus, f 9(x) 5 0 only when kak 5 0 for k 5 1, . . . , n. So, when char F 5 0, we have f(x) 5 a0, which is not an irreducible polynomial. This contradicts the hypothesis that f(x) is irreducible over F. Thus, f(x) has no multiple zeros. When char F 5 p 2 0, we have ak 5 0 when p does not divide k. Thus, the only powers of x that appear in the sum anx n 1 ? ? ? 1 a1x 1 a0 are those of the form x pj 5 (x p) j. It follows that f(x) 5 g(x p) for some g(x) [ F[x]. [For example, if f(x) 5 x4p 1 3x2p 1 x p 1 1, then g(x) 5 x4 1 3x2 1 x 1 1.] Theorem 20.6 shows that an irreducible polynomial over a field of characteristic 0 cannot have multiple zeros. The desire to extend this result to a larger class of fields motivates the following definition. Definition

A field F is called perfect if F has characteristic 0 or if F has characteristic p and Fp 5 {ap | a [ F} 5 F.

The most important family of perfect fields of characteristic p is the finite fields. Theorem 20.7 Finite Fields Are Perfect Every finite field is perfect.

PROOF Let F be a finite field of characteristic p. Consider the mapping f from F to F defined by f(x) 5 xp for all x [ F. We claim that f is a field automorphism. Obviously, f(ab) 5 (ab)p 5 a pb p 5 p f(a)f(b). Moreover, f(a 1 b) 5 (a 1 b)p 5 ap 1 a b ap21b 1 1 p p22 2 p a b a b 1???1 a b ab p21 1 b p 5 a p 1 b p, since each 2 p21 p a b is divisible by p. Finally, since xp 2 0 when x 2 0, Ker f 5 {0}. i Thus, f is one-to-one and, since F is finite, f is onto. This proves that Fp 5 F.

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Theorem 20.8 Criterion for No Multiple Zeros If f(x) is an irreducible polynomial over a perfect field F, then f(x) has no multiple zeros.

PROOF The case where F has characteristic 0 has been done. So let us assume that f (x) [ F[x] is irreducible over a perfect field F of characteristic p and that f (x) has multiple zeros. From Theorem 20.6 we know that f (x) 5 g(x p) for some g(x) [ F[x], say, g(x) 5 anx n 1 an21x n21 1 ? ? ? 1 a1x 1 a0. Since Fp 5 F, each ai in F can be written in the form bip for some bi in F. So, using Exercise 45a in Chapter 13, we have p

f(x) 5 g(x p) 5 bn px pn 1 bn21 px p(n21) 1 ? ? ? 1 b1 x p 1 b0

p

5 (bn x n 1 bn21x n21 1 ? ? ? 1 b1x 1 b0)p 5 (h(x))p where h(x) [ F[x]. But then f(x) is not irreducible. The next theorem shows that when an irreducible polynomial does have multiple zeros, there is something striking about the multiplicities. Theorem 20.9 Zeros of an Irreducible over a Splitting Field Let f(x) be an irreducible polynomial over a field F and let E be a splitting field of f(x) over F. Then all the zeros of f(x) in E have the same multiplicity.

PROOF Let a and b be distinct zeros of f(x) in E. If a has multiplicity m, then in E[x] we may write f(x) 5 (x 2 a)mg(x). It follows from the lemma preceding Theorem 20.4 and from Theorem 20.4 that there is a field isomorphism f from E to itself that carries a to b and acts as the identity on F. Thus, f(x) 5 f( f(x)) 5 (x 2 b)mf(g(x)) and we see that the multiplicity of b is greater than or equal to the multiplicity of a. By interchanging the roles of a and b, we observe that the multiplicity of a is greater than or equal to the multiplicity of b. So, we have proved that a and b have the same multiplicity. As an immediate corollary of Theorem 20.9 we have the following appealing result.

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Corollary Factorization of an Irreducible over a Splitting Field Let f(x) be an irreducible polynomial over a field F and let E be a splitting field of f(x). Then f(x) has the form a(x 2 a1)n(x 2 a2)n ? ? ? (x 2 at)n where a1, a2, . . . , at are distinct elements of E and a [ F.

We conclude this chapter by giving an example of an irreducible polynomial over a field that does have a multiple zero. In particular, notice that the field we use is not perfect. EXAMPLE 9 Let F 5 Z2(t) be the field of quotients of the ring Z2[t] of polynomials in the indeterminate t with coefficients from Z2. (We must introduce a letter other than x, since the members of F are going to be our coefficients for the elements in F[x].) Consider f(x) 5 x2 2 t [ F[x]. To see that f(x) is irreducible over F, it suffices to show that it has no zeros in F. Well, suppose that h(t)/k(t) is a zero of f(x). Then (h(t)/k(t))2 5 t, and therefore (h(t))2 5 t(k(t))2. Since h(t), k(t) [ Z2[t], we then have h(t2) 5 tk(t2) (see Exercise 45 in Chapter 13). But deg h(t2) is even, whereas deg tk(t2) is odd. So, f(x) is irreducible over F. Finally, since t is a constant in F[x] and the characteristic of F is 2, we have f9(x) 5 0, so that f9(x) and f(x) have f(x) as a common factor. So, by Theorem 20.5, f(x) has a multiple zero in some extension of F. (Indeed, it has a single zero of multiplicity 2 in K 5 F[x]/x2 2 t.)

Exercises I have yet to see any problem, however complicated, which, when you looked at it in the right way, did not become still more complicated. PAUL ANDERSON,

New Scientist

1. Describe the elements of Q(" 5). 2. Show that Q("2, "3) 5 Q("2 1 "3). 3. Find the splitting field of x3 2 1 over Q. Express your answer in the form Q(a). 4. Find the splitting field of x4 1 1 over Q. 5. Find the splitting field of 3

x4 1 x2 1 1 5 (x2 1 x 1 1)(x2 2 x 1 1) over Q.

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6. Let a, b [ R with b 2 0. Show that R(a 1 bi) 5 C. _ 7. Find a polynomial p(x) in Q[x] such that Q("11 " 5) is ringisomorphic to Q[x]/p(x). 8. Let F 5 Z2 and let f (x) 5 x3 1 x 1 1 [ F[x]. Suppose that a is a zero of f(x) in some extension of F. How many elements does F(a) have? Express each member of F(a) in terms of a. Write out a complete multiplication table for F(a). 9. Let F(a) be the field described in Exercise 8. Express each of a5, a22, and a100 in the form c2a2 1 c1a 1 c0. 10. Let F(a) be the field described in Exercise 8. Show that a2 and a2 1 a are zeros of x3 1 x 1 1. 11. Describe the elements in Q(p). 12. Let F 5 Q(p3). Find a basis for F(p) over F. 13. Write x7 2 x as a product of linear factors over Z3. Do the same for x10 2 x. 3 14. Find all ring automorphisms of Q( " 5). 15. Let F be a field of characteristic p and let f (x) 5 x p 2 a [ F[x]. Show that f(x) is irreducible over F or f(x) splits in F. 16. Suppose that b is a zero of f (x) 5 x 4 1 x 1 1 in some field extension E of Z2. Write f(x) as a product of linear factors in E[x]. 17. Find a, b, c in Q such that (1 1 "4)/(2 2 "2) 5 a 1 b "2 1 c " 4. 3

3

3

3

Note that such a, b, c exist, since (1 1 "4)/(2 2 "2) [ Q("2) 5 {a 1 b"2 1 c"4 | a, b, c [ Q}. 3

3

3

3

3

18. Express (3 1 4"2)21 in the form a 1 b"2, where a, b [ Q. 19. Show that Q(4 2 i) 5 Q(1 1 i), where i 5 "21. 20. Let F be a field, and let a and b belong to F with a 2 0. If c belongs to some extension of F, prove that F(c) 5 F(ac 1 b). (F “absorbs” its own elements.) 21. Let f(x) [ F[x] and let a [ F. Show that f(x) and f(x 1 a) have the same splitting field over F. 22. Recall that two polynomials f (x) and g(x) from F[x] are said to be relatively prime if there is no polynomial of positive degree in F[x] that divides both f (x) and g(x). Show that if f (x) and g(x) are relatively prime in F[x], they are relatively prime in K[x], where K is any extension of F. 23. Determine all of the subfields of Q("2).

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Fields

24. Let E be an extension of F and let a and b belong to E. Prove that F(a, b) 5 F(a)(b) 5 F(b)(a). 25. Write x3 1 2x 1 1 as a product of linear polynomials over some field extension of Z3. 26. Express x8 2 x as a product of irreducibles over Z2. 27. Prove or disprove that Q("3) and Q("2 3) are ring-isomorphic. 28. For any prime p, find a field of characteristic p that is not perfect. 29. If b is a zero of x2 1 x 1 2 over Z5, find the other zero. 30. Show that x4 1 x 1 1 over Z2 does not have any multiple zeros in any field extension of Z2. 31. Show that x21 1 2x8 1 1 does not have multiple zeros in any extension of Z3. 32. Show that x21 1 2x9 1 1 has multiple zeros in some extension of Z3. 33. Let F be a field of characteristic p 2 0. Show that the polynomial n f(x) 5 x p 2 x over F has distinct zeros. 34. Find the splitting field for f (x) 5 (x2 1 x 1 2)(x2 1 2x 1 2) over Z3[x]. Write f(x) as a product of linear factors. 35. Let F, K, and L be fields with F # K # L. If L is a splitting field for some nonconstant polynomial f(x) over F, show that L is a splitting field for f(x) over K. 36. If a ? 0 belongs to a field F and x n 2 a splits in some extension E of F, prove that E contains all the nth roots of unity. 37. Suppose that f(x) is a fifth-degree polynomial that is irreducible over Z2 . Prove that every nonidentity element is a generator of the cyclic group (Z2 3x4>f(x) ) *.

Leopold Kronecker He [Kronecker] wove together the three strands of his greatest interests—the theory of numbers, the theory of equations and elliptic functions—into one beautiful pattern. E. T. BELL

LEOPOLD KRONECKER was born on December 7, 1823, in Leignitz, Prussia. As a schoolboy, he received special instruction from the great algebraist Kummer. Kronecker entered the University of Berlin in 1841 and completed his Ph.D. dissertation in 1845 on the units in a certain ring. Kronecker devoted the years 1845–1853 to business affairs, relegating mathematics to a hobby. Thereafter, being well-off financially, he spent most of his time doing research in algebra and number theory. Kronecker was one of the early advocates of the abstract approach to algebra. He innovatively applied rings and fields in his investigations of algebraic numbers, established the Fundamental Theorem of Finite Abelian Groups, and was the first mathematician to master Galois’s theory of fields.

Kronecker advocated constructive methods for all proofs and definitions. He believed that all mathematics should be based on relationships among integers. He went so far as to say to Lindemann, who proved that p is transcendental, that irrational numbers do not exist. His most famous remark on the matter was “God made the integers, all the rest is the work of man.” Henri Poincaré once remarked that Kronecker was able to produce fine work in number theory and algebra only by temporarily forgetting his own philosophy. Kronecker died on December 29, 1891, at the age of 68. For more information about Kronecker, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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21 Algebraic Extensions

Banach once told me, “Good mathematicians see analogies between theorems or theories, the very best ones see analogies between analogies.” S. M. ULAM, Adventures of a Mathematician

Characterization of Extensions In Chapter 20, we saw that every element in the field Q("2) has the particularly simple form a 1 b"2, where a and b are rational. On the other hand, the elements of Q(p) have the more complicated form (anpn 1 an21pn21 1 ? ? ? 1 a0)/(bmpm 1 bm21pm21 1 ? ? ? 1 b0), where the a’s and b’s are rational. The fields of the first type have a great deal of structure. This structure is the subject of this chapter. Definition Types of Extensions

Let E be an extension field of a field F and let a [ E. We call a algebraic over F if a is the zero of some nonzero polynomial in F[x]. If a is not algebraic over F, it is called transcendental over F. An extension E of F is called an algebraic extension of F if every element of E is algebraic over F. If E is not an algebraic extension of F, it is called a transcendental extension of F. An extension of F of the form F(a) is called a simple extension of F.

Leonhard Euler used the term transcendental for numbers that are not algebraic because “they transcended the power of algebraic methods.” Although Euler made this distinction in 1744, it wasn’t until 1844 that the existence of transcendental numbers over Q was proved by Joseph Liouville. Charles Hermite proved that e is transcendental over Q in 1873, and Lindemann showed that p is transcendental over Q in 1882. To this day, it is not known whether p 1 e is transcendental over Q. With a precise definition of “almost all,” it can be shown that almost all real numbers are transcendental over Q.

370

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Theorem 21.1 shows why we make the distinction between elements that are algebraic over a field and elements that are transcendental over a field. Recall that F(x) is the field of quotients of F[x]; that is, F(x) 5 {f(x)/g(x) | f(x), g(x) [ F[x], g(x) 2 0}. Theorem 21.1 Characterization of Extensions Let E be an extension field of the field F and let a [ E. If a is transcendental over F, then F(a) < F(x). If a is algebraic over F, then F(a) < F[x]/ p(x), where p(x) is a polynomial in F[x] of minimum degree such that p(a) 5 0. Moreover, p(x) is irreducible over F.

PROOF Consider the homomorphism f:F[x] S F(a) given by f (x) S f (a). If a is transcendental over F, then Ker f 5 {0}, and so we may extend f to an isomorphism f:F(x) S F(a) by defining f ( f(x)/g(x)) 5 f(a)/g(a). If a is algebraic over F, then Ker f 2 {0}; and, by Theorem 16.4, there is a polynomial p(x) in F[x] such that Ker f 5 p(x) and p(x) has minimum degree among all nonzero elements of Ker f. Thus, p(a) 5 0 and, since p(x) is a polynomial of minimum degree with this property, it is irreducible over F. The proof of Theorem 21.1 can readily be adapted to yield the next two results also. The details are left to the reader (see Exercise 1). Theorem 21.2 Uniqueness Property If a is algebraic over a field F, then there is a unique monic irreducible polynomial p(x) in F[x] such that p(a) 5 0.

The polynomial with the property specified in Theorem 21.2 is called the minimal polynomial for a over F. Theorem 21.3 Divisibility Property Let a be algebraic over F, and let p(x) be the minimal polynomial for a over F. If f(x) [ F[x] and f(a) 5 0, then p(x) divides f(x) in F[x].

If E is an extension field of F, we may view E as a vector space over F (that is, the elements of E are the vectors and the elements of F are the scalars). We are then able to use such notions as dimension and basis in our discussion.

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Fields

Finite Extensions Definition Degree of an Extension

Let E be an extension field of a field F. We say that E has degree n over F and write [E:F] 5 n if E has dimension n as a vector space over F. If [E:F] is finite, E is called a finite extension of F; otherwise, we say that E is an infinite extension of F.

Figure 21.1 illustrates a convenient method of depicting the degree of a field extension over a field. Q(√2)

3

Q(√2)

3

6

2 Q [Q(√2):Q] = 2

6

Q(√2)

Q

Q

3

[Q(√2):Q] = 3

6

[Q(√2):Q] = 6

E

n F [E:F] = n

Figure 21.1

EXAMPLE 1 The field of complex numbers has degree 2 over the reals, since {1, i} is a basis. The field of complex numbers is an infinite extension of the rationals. EXAMPLE 2 If a is algebraic over F and its minimal polynomial over F has degree n, then, by Theorem 20.3, we know that {1, a, . . . , an21} is a basis for F(a) over F; and, therefore, [F(a):F] 5 n. In this case, we say that a has degree n over F. Theorem 21.4 Finite Implies Algebraic If E is a finite extension of F, then E is an algebraic extension of F.

PROOF Suppose that [E:F] 5 n and a [ E. Then the set {1, a, . . . , an} is linearly dependent over F; that is, there are elements c0, c1, . . . , cn in F, not all zero, such that cnan 1 cn21an21 1 ? ? ? 1 c1a 1 c0 5 0. Clearly, then, a is a zero of the nonzero polynomial f(x) 5 cnx n 1 cn21x n21 1 ? ? ? 1 c1x 1 c0.

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The converse of Theorem 21.4 is not true, for otherwise, the degrees of the elements of every algebraic extension of E over F would 3 4 be bounded. But Q("2, " 2, " 2, . . .) is an algebraic extension of Q that contains elements of every degree over Q (see Exercise 3). The next theorem is the field theory counterpart of Lagrange’s Theorem for finite groups. Like all counting theorems, it has far-reaching consequences. Theorem 21.5 [K:F] 5 [K:E][E:F] Let K be a finite extension field of the field E and let E be a finite extension field of the field F. Then K is a finite extension field of F and [K:F] 5 [K:E][E:F].

PROOF Let X 5 {x1, x2, . . . , xn} be a basis for K over E, and let Y 5 {y1, y2, . . . , ym} be a basis for E over F. It suffices to prove that YX 5 {yj xi | 1 # j # m, 1 # i # n} is a basis for K over F. To do this, let a [ K. Then there are elements b1, b2, . . . , bn [ E such that a 5 b1x1 1 b2x2 1 ? ? ? 1 bnxn. And, for each i 5 1, . . . , n, there are elements ci1, ci2, . . . , cim [ F such that bi 5 ci1y1 1 ci2 y2 1 ? ? ? 1 cimym. Thus,

n

n

m

i51

i51

j51

a 5 a bi xi 5 a a a cij yj b xi 5 a cij (yj xi ). i, j

This proves that YX spans K over F. Now suppose there are elements cij in F such that

0 5 g c ij (yj x i) 5 g A g (c ij yj)Bx i. i,j

i

j

Then, since each g c ij yj [ E and X is a basis for K over E, we have j

g c ij yj 5 0 j

for each i. But each cij [ F and Y is a basis for E over F, so each cij 5 0. This proves that the set YX is linearly independent over F. Using the fact that for any field extension L of a field J, [L:J] 5 n if and only if L is isomorphic to Jn as vector spaces (see Exercise 29), we may give a concise conceptual proof of Theorem 21.5, as follows. Let

374

Fields K n nm

E m F

[K:F ] = [K:E ][E:F ]

Figure 21.2

[K:E] 5 n and [E:F] 5 m. Then K < En and E < Fm, so that K < En < (Fm)n < Fmn. Thus, [K:F] 5 mn. The content of Theorem 21.5 can be pictured as in Figure 21.2. Examples 3, 4, and 5 show how Theorem 21.5 is often utilized. EXAMPLE 3 Since {1, "3} is a basis for Q("3, "5) over Q("5) (see Exercise 7) and {1, "5} is a basis for Q("5) over Q, the proof of Theorem 21.5 shows that {1, "3, "5, "15} is a basis for Q("3, "5) over Q. (See Figure 21.3.) EXAMPLE 4 Consider Q(" 2, " 3). Then [Q(" 2, " 3) : Q] 5 12. 3 4 3 4 3 3 For, clearly, [Q(" 2, " 3) : Q] 5 [Q(" 2, " 3) : Q(" 2)] [Q(" 2) : Q] 3 4 3 4 4 4 and [Q(" 2, " 3) : Q] 5 [Q(" 2, " 3) : Q(" 3)] [Q(" 3) : Q] show that 3 4 3 4 both 3 5 [Q(" 2) : Q] and 4 5 [Q(" 3) : Q] divide [Q(" 2, " 3) : Q]. 3 4 3 4 3 Thus, [Q(" 2, " 3) : Q] $ 12. On the other hand, [Q(" 2, " 3) : Q(" 2)] 4 3 is at most 4, since " 3 is a zero of x4 2 3 [ Q(" 2)[x]. Therefore, 3 4 3 4 3 3 [Q(" 2, " 3) : Q] 5 [Q(" 2, " 3 ): Q(" 2)][Q(" 2) : Q] # 4 ? 3 5 12. (See Figure 21.4.) 3

4

3

3

Q(√3,√5 ) 2

4 Q(√5)

4 2

4

Q(√2, √3) 2

Q(√3)

4

2 Q

Figure 21.3

3

3

Q(√2)

4

Q(√3)

12 3

4 Q

Figure 21.4

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375

Theorem 21.5 can sometimes be used to show that a field does not contain a particular element. EXAMPLE 5 Recall from Example 7 in Chapter 17 that h(x) 5 15x4 2 10x2 1 9x 1 21 is irreducible over Q. Let b be a zero of h(x) in some extension of Q. Then, even though we don’t know what b is, we can still 3 3 prove that " 2 is not an element of Q(b). For, if so, then Q , Q(" 2) # 3 3 Q(b) and 4 5 [Q(b) : Q] 5 [Q(b) : Q(" 2)][Q(" 2) : Q] implies that 3 divides 4. Notice that this argument cannot be used to show that "2 is not contained in Q(b). EXAMPLE 6 Consider Q("3, "5). We claim that Q("3, "5) 5 Q("3 1 "5). The inclusion Q("3 1 "5) # Q("3, "5) is clear. 1 "3 2 "5 Now note that since ("3 1 "5)21 5 5 ? "3 1 "5 "3 2 "5 1 2 ("3 2 "5), we know that "3 2 "5 belongs to Q ("3 1 "5). It 2 follows that [("3 1 "5) 1 ("3 2 "5)]/2 5 "3 and [("3 1 "5 )2 ("3 2 "5)]/2 5 "5 both belong to Q("3 1 "5), and therefore Q("3, "5) # Q("3 1 "5 ). EXAMPLE 7 It follows from Example 6 and Theorem 20.3 that the minimal polynomial for "3 1 "5 over Q has degree 4. How can we find this polynomial? We begin with x 5 "3 1 "5. Then x2 5 3 1 2"15 1 5. From this we obtain x2 2 8 5 2"15 and, by squaring both sides, x4 2 16x 1 64 5 60. Thus, "3 1 "5 is a zero of x4 2 16x 1 4. We know that this is the minimal polynomial of "3 1 "5 over Q since it is monic and has degree 4. Example 6 shows that an extension obtained by adjoining two elements to a field can sometimes be obtained by adjoining a single element to the field. Our next theorem shows that, under certain conditions, this can always be done. Theorem 21.6 Primitive Element Theorem (Steinitz, 1910) If F is a field of characteristic 0, and a and b are algebraic over F, then there is an element c in F(a, b) such that F(a, b) 5 F(c).

PROOF Let p(x) and q(x) be the minimal polynomials over F for a and b, respectively. In some extension K of F, let a1, a2, . . . , am and b1, b2, . . . , bn be the distinct zeros of p(x) and q(x), respectively, where a 5 a1

376

Fields

and b 5 b1. Among the infinitely many elements of F, choose an element d not equal to (ai 2 a)/(b 2 bj) for all i $ 1 and all j . 1. In particular, ai 2 a 1 d(b 2 bj) for j . 1. We shall show that c 5 a 1 db has the property that F(a, b) 5 F(c). Certainly, F(c) # F(a, b). To verify that F(a, b) # F(c), it suffices to prove that b [ F(c), for then b, c, and d belong to F(c) and a 5 c 2 bd. Consider the polynomials q(x) and r (x) 5 p(c 2 dx) [that is, r (x) is obtained by substituting c 2 dx for x in p(x)] over F(c). Since both q(b) 5 0 and r(b) 5 p(c 2 db) 5 p(a) 5 0, both q(x) and r(x) are divisible by the minimal polynomial s(x) for b over F(c) (see Theorem 21.3). Because s(x) [ F(c)[x], we may complete the proof by proving that s(x) 5 x 2 b. Since s(x) is a common divisor of q(x) and r(x), the only possible zeros of s(x) in K are the zeros of q(x) that are also zeros of r (x). But r(bj) 5 p(c 2 dbj) 5 p(a 1db 2 dbj) 5 p(a 1 d(b 2 bj)) and d was chosen such that a 1 d(b 2 bj) 2 ai for j . 1. It follows that b is the only zero of s(x) in K[x] and, therefore, s(x) 5 (x 2 b)u. Since s(x) is irreducible and F has characteristic 0, Theorem 20.6 guarantees that u 5 1. In the terminology introduced earlier, it follows from Theorem 21.6 and induction that any finite extension of a field of characteristic 0 is a simple extension. An element a with the property that E 5 F(a) is called a primitive element of E.

Properties of Algebraic Extensions Theorem 21.7 Algebraic over Algebraic Is Algebraic If K is an algebraic extension of E and E is an algebraic extension of F, then K is an algebraic extension of F.

PROOF Let a [ K. It suffices to show that a belongs to some finite extension of F. Since a is algebraic over E, we know that a is the zero of some irreducible polynomial in E[x], say, p(x) 5 bnx n 1 ? ? ? 1 b0. Now we construct a tower of field extensions of F, as follows: F0 5 F(b0), F1 5 F0(b1), . . . , Fn 5 Fn21(bn). In particular, Fn 5 F(b0, b1, . . . , bn),

21 | Algebraic Extensions

377

K E

Fn(a) Fn Fn21 F1 F0 F

Figure 21.5

so that p(x) [ Fn[x]. Thus, [Fn(a):Fn] 5 n; and, because each bi is algebraic over F, we know that each [Fi11:Fi] is finite. So, [Fn(a):F ] 5 [Fn(a):Fn][Fn:Fn21] ? ? ? [F1:F0][F0:F] is finite. (See Figure 21.5.) Corollary Subfield of Algebraic Elements Let E be an extension field of the field F. Then the set of all elements of E that are algebraic over F is a subfield of E.

PROOF Suppose that a, b [ E are algebraic over F and b 2 0. To show that a 1 b, a 2 b, ab, and a/b are algebraic over F, it suffices to show that [F(a, b):F] is finite, since each of these four elements belongs to F(a, b). But note that [F(a, b):F] 5 [F(a, b):F(b)][F(b):F]. Also, since a is algebraic over F, it is certainly algebraic over F(b). Thus, both [F(a, b):F(b)] and [F(b):F] are finite. For any extension E of a field F, the subfield of E of the elements that are algebraic over F is called the algebraic closure of F in E. One might wonder if there is such a thing as a maximal algebraic extension of a field F—that is, whether there is an algebraic extension E of F that has no proper algebraic extensions. For such an E to exist, it is necessary that every polynomial in E[x] splits in E. Otherwise, it follows from Kronecker’s Theorem that E would have a proper algebraic extension. This condition is also sufficient. If every member of E[x] splits in E, and K is an algebraic extension of E, then every member of K is a zero of

378

Fields

some element of E[x]. But the zeros of elements of E[x] are in E. A field that has no proper algebraic extension is called algebraically closed. In 1910, Ernst Steinitz proved that every field F has a unique (up to isomorphism) algebraic extension that is algebraically closed. This field is called the algebraic closure of F. A proof of this result requires a sophisticated set theory background. In 1799, Gauss, at the age of 22, proved that C is algebraically closed. This fact was considered so important at the time that it was called “The Fundamental Theorem of Algebra.” Over a 50-year period, Gauss found three additional proofs of the Fundamental Theorem. Today more than 100 proofs exist. In view of the ascendancy of abstract algebra in the 20th century, a more appropriate phrase for Gauss’s result would be “The Fundamental Theorem of Classical Algebra.”

Exercises It matters not what goal you seek Its secret here reposes: You’ve got to dig from week to week To get Results or Roses. EDGAR GUEST

1. Prove Theorem 21.2 and Theorem 21.3. 2. Let E be the algebraic closure of F. Show that every polynomial in F[x] splits in E. 3 4 3. Prove that Q("2, " 2, " 2, . . .) is an algebraic extension of Q but not a finite extension of Q. (This exercise is referred to in this chapter.) 4. Let E be an algebraic extension of F. If every polynomial in F[x] splits in E, show that E is algebraically closed. 5. Suppose that F is a field and every irreducible polynomial in F[x] is linear. Show that F is algebraically closed. 6. Suppose that f (x) and g(x) are irreducible over F and that deg f (x) and deg g(x) are relatively prime. If a is a zero of f (x) in some extension of F, show that g(x) is irreducible over F(a). 7. Let a and b belong to Q with b 2 0. Show that Q("a) 5 Q("b) if and only if there exists some c [ Q such that a 5 bc2. 8. Find the degree and a basis for Q("3 1 "5) over Q("15). Find 3 4 the degree and a basis for Q("2, " 2, " 2) over Q. 9. Suppose that E is an extension of F of prime degree. Show that, for every a in E, F(a) 5 F or F(a) 5 E.

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10. Let a be a complex number that is algebraic over Q. Does your n argument work when "a is replaced by "a? Show that "a is algebraic over Q. 11. Suppose that E is an extension of F and a, b [ E. If a is algebraic over F of degree m, and b is algebraic over F of degree n, where m and n are relatively prime, show that [F(a, b):F] 5 mn. 12. Find an example of a field F and elements a and b from some extension field such that F(a, b) 2 F(a), F(a, b) 2 F(b), and [F(a, b):F] , [F(a):F][F(b):F]. 13. Let K be a field extension of F and let a [ K. Show that [F(a):F(a3)] # 3. Find examples to illustrate that [F(a):F(a3)] can be 1, 2, or 3. 14. Find the minimal polynomial for "2 3 1 "2 over Q. 15. Let K be an extension of F. Suppose that E1 and E2 are contained in K and are extensions of F. If [E1:F] and [E2:F] are both prime, show that E1 5 E2 or E1 > E2 5 F. 3 3 16. Find the minimal polynomial for " 2 1 " 4 over Q. 17. Let E be a finite extension of R. Use the fact that C is algebraically closed to prove that E 5 C or E 5 R. 18. Suppose that [E:Q] 5 2. Show that there is an integer d such that E 5 Q("d) where d is not divisible by the square of any prime. 19. Suppose that p(x) [ F[x] and E is a finite extension of F. If p(x) is irreducible over F and deg p(x) and [E:F] are relatively prime, show that p(x) is irreducible over E. 20. Let E be a field extension of F. Show that [E:F] is finite if and only if E 5 F(a1, a2, . . . , an), where a1, a2, . . . , an are algebraic over F. 21. If a and b are real numbers and a and b are transcendental over Q, show that either ab or a 1 b is also transcendental over Q. 22. Let f (x) be a nonconstant element of F[x]. If a belongs to some extension of F and f (a) is algebraic over F, prove that a is algebraic over F. 23. Let f (x) 5 ax2 1 bx 1 c [ Q[x]. Find a primitive element for the splitting field for f(x) over Q. 24. Find the splitting field for x4 2 x2 2 2 over Z3. 25. Let f (x) [ F[x]. If deg f (x) 5 2 and a is a zero of f (x) in some extension of F, prove that F(a) is the splitting field for f(x) over F. 26. Let a be a complex zero of x2 1 x 1 1 over Q. Prove that Q("a) 5 Q(a). 27. If F is a field and the multiplicative group of nonzero elements of F is cyclic, prove that F is finite.

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Fields

28. Let a be a complex number that is algebraic over Q and let r be a rational number. Show that ar is algebraic over Q. 29. Prove that, if K is a field extension of F, then [K:F] 5 n if and only if K is isomorphic to Fn as vector spaces. (See Exercise 27 in Chapter 19 for the appropriate definition. This exercise is referred to in this chapter.) 30. Let a be a positive real number and let n be an integer greater than 1. Prove or disprove that [Q(a1/n):Q] 5 n. 31. Let a and b belong to some extension of F and let b be algebraic over F. Prove that [F(a, b):F(a)] # [F(a, b):F]. 32. Let f (x) and g(x) be irreducible polynomials over a field F and let a and b belong to some extension E of F. If a is a zero of f (x) and b is a zero of g(x), show that f (x) is irreducible over F(b) if and only if g(x) is irreducible over F(a). 33. Let b be a zero of f(x) 5 x5 1 2x 1 4 (see Example 8 in Chapter 17). 3 4 Show that none of "2, " 2, " 2 belongs to Q(b). 3 6 34. Prove that Q("2, " 2) 5 Q(" 2). 35. Let a and b be rational numbers. Show that Q("a, "b) 5 Q("a1"b). 36. Let F, K and L be fields with F 8 K 8 L. If L is a finite extension of F and 3L : F4 5 3L : K4 , prove that F 5 K . 37. Let F be a field and K a splitting field for some nonconstant polynomial over F. Show that K is a finite extension of F. 38. Prove that C is not the splitting field of any polynomial in Q3x4 . 39. Prove that "2 is not an element of Q(p) . 2p 2p 2p 40. Let a 5 cos 2p 7 1 i sin 7 and b 5 cos 5 1 i sin 5 . Prove that b is not in Q(a) . 41. Suppose that a is algebraic over a field F. Show that a and 1 1 a21 have the same degree over F.

Suggested Readings R. L. Roth, “On Extensions of Q by Square Roots,” American Mathematical Monthly 78 (1971): 392–393. In this paper, it is proved that if p1, p2, . . . , pn are distinct primes, then [Q(" p1," p2, . . . , " pn):Q] 5 2n. Paul B. Yale, “Automorphisms of the Complex Numbers,” Mathematics Magazine 39 (1966): 135–141. This award-winning expository paper is devoted to various results on automorphisms of the complex numbers.

Irving Kaplansky He got to the top of the heap by being a first-rate doer and expositor of algebra. PAUL R. HALMOS, I Have a

Photographic Memory

IRVING KAPLANSKY was born on March 22, 1917, in Toronto, Canada, a few years after his parents emigrated from Poland. Although his parents thought he would pursue a career in music, Kaplansky knew early on that mathematics was what he wanted to do. As an undergraduate at the University of Toronto, Kaplansky was a member of the winning team in the first William Lowell Putnam competition, a mathematical contest for United States and Canadian college students. Kaplansky received a B.A. degree from Toronto in 1938 and an M.A. in 1939. In 1939, he entered Harvard University to earn his doctorate as the first recipient of a Putnam Fellowship. After receiving his Ph.D. from Harvard in 1941, Kaplansky stayed on as Benjamin Peirce instructor until 1944. After one year at Columbia University, he went to the University of Chicago, where he remained until his retirement in 1984. He then became the director of the Mathematical Sciences Research Institute at the University of California, Berkeley.

Kaplansky’s interests were broad, including areas such as ring theory, group theory, field theory, Galois theory, ergodic theory, algebras, metric spaces, number theory, statistics, and probability. Among the many honors Kaplansky received are election to both the National Academy of Sciences and the American Academy of Arts and Sciences, election to the presidency of the American Mathematical Society, and the 1989 Steele Prize for cumulative influence from the American Mathematical Society. The Steele Prize citation says, in part, “. . . he has made striking changes in mathematics and has inspired generations of younger mathematicians.” Kaplansky died on June 25, 2006 at the age of 89. For more information about Kaplansky, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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22 Finite Fields

This theory [of finite fields] is of considerable interest in its own right and it provides a particularly beautiful example of how the general theory of the preceding chapters fits together to provide a rather detailed description of all finite fields. RICHARD A. DEAN, Elements of Abstract Algebra

Classification of Finite Fields In this, our final chapter on field theory, we take up one of the most beautiful and important areas of abstract algebra—finite fields. Finite fields were first introduced by Galois in 1830 in his proof of the unsolvability of the general quintic equation. When Cayley invented matrices a few decades later, it was natural to investigate groups of matrices over finite fields. To this day, matrix groups over finite fields are among the most important classes of groups. In the past 50 years, there have been important applications of finite fields in computer science, coding theory, information theory, and cryptography. But, besides the many uses of finite fields in pure and applied mathematics, there is yet another good reason for studying them. They are just plain fun! The most striking fact about finite fields is the restricted nature of their order and structure. We have already seen that every finite field has prime-power order (Exercise 47 in Chapter 13). A converse of sorts is also true. Theorem 22.1 Classification of Finite Fields For each prime p and each positive integer n, there is, up to isomorphism, a unique finite field of order p n.

PROOF Consider the splitting field E of f(x) 5 xpn 2 x over Zp. We will show that |E| 5 pn. Since f(x) splits in E, we know that f(x) has exactly pn zeros in E, counting multiplicity. Moreover, by Theorem 20.5, every zero of f(x) has multiplicity 1. Thus, f(x) has pn distinct zeros in E. 382

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On the other hand, the set of zeros of f(x) in E is closed under addition, subtraction, multiplication, and division by nonzero elements (see Exercise 35), so that the set of zeros of f(x) is itself a field extension of Zp in which f(x) splits. Thus, the set of zeros of f(x) is E and, therefore, |E| 5 pn. To show that there is a unique field for each prime-power, suppose that K is any field of order pn. Then K has a subfield isomorphic to Zp (generated by 1), and, because the nonzero elements of K form a multiplicative group of order pn 2 1, every element of K is a zero of f(x) 5 n xp 2 x (see Exercise 25). So, K must be a splitting field for f(x) over Zp. By the corollary to Theorem 20.4, there is only one such field up to isomorphism. The existence portion of Theorem 22.1 appeared in the works of Galois and Gauss in the first third of the 19th century. Rigorous proofs were given by Dedekind in 1857 and by Jordan in 1870 in his classic book on group theory. The uniqueness portion of the theorem was proved by E. H. Moore in an 1893 paper concerning finite groups. The mathematics historian E. T. Bell once said that this paper by Moore marked the beginning of abstract algebra in America. Because there is only one field for each prime-power pn, we may unambiguously denote it by GF( pn), in honor of Galois, and call it the Galois field of order pn.

Structure of Finite Fields The next theorem tells us the additive and multiplicative group structure of a field of order pn. Theorem 22.2 Structure of Finite Fields As a group under addition, GF(pn) is isomorphic to Zp % Zp % ? ? ? % Zp. n factors As a group under multiplication, the set of nonzero elements of GF( pn) is isomorphic to Zpn21 (and is, therefore, cyclic).

PROOF Since GF( pn) has characteristic p (Theorem 13.3), every nonzero element of GF( pn) has additive order p. Then by the Fundamental Theorem of Finite Abelian Groups, GF( pn) under addition is isomorphic to a direct product of n copies of Zp.

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Fields

To see that the multiplicative group GF( pn)* of nonzero elements of GF(pn) is cyclic, we first note that by the Fundamental Theorem of Abelian Groups (Theorem 11.1), GF(pn)* is isomorphic to a direct product of the form Z n1 % Z n2 % ? ? ? % Z nm. If the orders of these components are pairwise relatively prime then it follows from Corollary 1 of Theorem 8.2 that GF( pn)* is cyclic. Hence we may assume that there is an integer d . 1 that divides the orders of two of the components. From the Fundamental Theorem of Cyclic Groups (Theorem 4.3) we know that each of these components has a subgroup of order d. This means that GF(pn)* has two distinct subgroups of order d, call them H and K. But then every element of H and K is a zero of xd 2 1, which contradicts the fact that a polynomial of degree d over a field can have at most d zeros (Corollary 3 of Theorem 16.2). Since Zp % Zp % ? ? ? % Zp is a vector space over Zp with {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1)} as a basis, we have the following useful and aesthetically appealing formula. Corollary 1 [GF(p n):GF(p)] 5 n

Corollary 2 GF(pn) Contains an Element of Degree n Let a be a generator of the group of nonzero elements of GF( pn) under multiplication. Then a is algebraic over GF( p) of degree n.

PROOF Observe that [GF( p)(a):GF( p)] 5 [GF( pn):GF( p)] 5 n. EXAMPLE 1 Let’s examine the field GF(16) in detail. Since x4 1 x 1 1 is irreducible over Z2, we know that GF(16) < {ax3 1 bx2 1 cx 1 d 1 x4 1 x 1 1 | a, b, c, d [ Z2}. Thus, we may think of GF(16) as the set F 5 {ax3 1 bx2 1 cx 1 d | a, b, c, d [ Z2}, where addition is done as in Z2[x], but multiplication is done modulo x4 1 x 1 1. For example, (x3 1 x2 1 x 1 1)(x3 1 x) 5 x3 1 x2,

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since the remainder upon dividing (x3 1 x2 1 x 1 1)(x3 1 x) 5 x6 1 x5 1 x2 1 x by x4 1 x 1 1 in Z2[x] is x3 1 x2. An easier way to perform the same calculation is to observe that in this context x4 1 x 1 1 is 0, so x4 5 2x 2 1 5 x 1 1, x5 5 x2 1 x, and x6 5 x3 1 x2. Thus, x6 1 x5 1 x2 1 x 5 (x3 1 x2) 1 (x2 1 x) 1 x2 1 x 5 x3 1 x2. Another way to simplify the multiplication process is to make use of the fact that the nonzero elements of GF(16) form a cyclic group of order 15. To take advantage of this, we must first find a generator of this group. Since any element F* must have a multiplicative order that divides 15, all we need to do is find an element a in F* such that a3 2 1 and a5 2 1. Obviously, x has these properties. So, we may think of GF(16) as the set {0, 1, x, x2, . . . , x14}, where x15 5 1. This makes multiplication in F trivial, but, unfortunately, it makes addition more difficult. For example, x10 ? x7 5 x17 5 x2, but what is x10 1 x7? So, we face a dilemma. If we write the elements of F* in the additive form ax3 1 bx2 1 cx 1 d, then addition is easy and multiplication is hard. On the other hand, if we write the elements of F* in the multiplicative form xi, then multiplication is easy and addition is hard. Can we have the best of both? Yes, we can. All we need to do is use the relation x4 5 x 1 1 to make a two-way conversion table, as in Table 22.1. So, we see from Table 22.1 that x10 1 x7 5 (x2 1 x 1 1) 1 (x3 1 x 1 1) 5 x3 1 x2 5 x6 and (x3 1 x2 1 1)(x3 1 x2 1 x 1 1) 5 x13 ? x12 5 x25 5 x10 5 x2 1 x 1 1. Don’t be misled by the preceding example into believing that the element x is always a generator for the cyclic multiplicative group of nonzero elements. It is not. (See Exercise 17.) Although any two

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Fields

Table 22.1 Conversion Table for Addition and Multiplication in GF(16) Multiplicative Form to Additive Form 1 x x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 x12 x13 x14

Additive Form to Multiplicative Form

1 x x2 x3 x11 x2 1 x x3 1 x2 x3 1 x 1 1 x2 1 1 x3 1 x x2 1 x 1 1 x3 1 x2 1 x x3 1 x2 1 x 1 1 x3 1 x2 1 1 x3 1 1

1 x x11 x2 x2 1 x x2 1 1 x2 1 x 1 1 x3 x3 1 x2 x3 1 x x3 1 1 x3 1 x2 1 x x3 1 x2 1 1 x3 1 x 1 1 x3 1 x2 1 x 1 1

1 x x4 x2 x5 x8 x10 x3 x6 x9 x14 x11 x13 x7 x12

irreducible polynomials of the same degree over Zp[x] yield isomorphic fields, some are better than others for computational purposes. EXAMPLE 2 Consider f(x) 5 x3 1 x2 1 1 over Z2. We will show how to write f(x) as the product of linear factors. Let F 5 Z2[x]/ f(x) and let a be a zero of f(x) in F. Then |F| 5 8 and |F*| 5 7. So, by Corollary 2 to Theorem 7.1, we know that |a| 5 7. Thus, by Theorem 20.3, F 5 {0, 1, a, a2, a3, a4, a5, a6} 5 {0, 1, a, a 1 1, a2, a2 1 a 1 1, a2 1 1, a2 1 a}. We know that a is one zero of f(x), and we can test the other elements of F to see if they are zeros. We can simplify the calculations by using the fact that a3 1 a2 1 1 5 0 to make a conversion table for the two forms of writing the elements of F. Because char F 5 2, we know that a3 5 a2 1 1. Then, a4 5 a3 1 a 5 (a2 1 1) 1 a 5 a2 1 a 1 1, a5 5 a3 1 a2 1 a 5 (a2 1 1) 1 a2 1 a 5 a 1 1, a6 5 a2 1 a, a7 5 1. Now let’s see whether a2 is a zero of f(x). f(a2) 5 (a2)3 1 (a2)2 1 1 5 a6 1 a4 1 1 5 (a2 1 a) 1 (a2 1 a 1 1) 1 1 5 0.

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So, yes, it is. Next we try a3. f(a3) 5 (a3)3 1 (a3)2 1 1 5 a9 1 a6 1 1 5 a2 1 (a2 1 a) 1 1 5 a 1 1 2 0. Now a4. f(a4) 5 (a4)3 1 (a4)2 1 1 5 a12 1 a8 1 1 5 a5 1 a 1 1 5 (a 1 1) 1 a 1 1 5 0. So, a4 is our remaining zero. Thus, f(x) 5 (x 2 a)(x 2 a2)(x 2 a4) 5 (x 1 a)(x 1 a2)(x 1 a4), since char F 5 2.

Subfields of a Finite Field Theorem 22.1 gives us a complete description of all finite fields. The following theorem gives us a complete description of all the subfields of a finite field. Notice the close analogy between this theorem and Theorem 4.3, which describes all the subgroups of a finite cyclic group. Theorem 22.3 Subfields of a Finite Field For each divisor m of n, GF( pn) has a unique subfield of order pm. Moreover, these are the only subfields of GF( pn).

PROOF To show the existence portion of the theorem, suppose that m divides n. Then, since pn 2 1 5 ( pm 2 1)( pn2m 1 pn22m 1 ? ? ? 1 pm 1 1), we see that pm 2 1 divides pn 2 1. For simplicity, write pn 2 1 5 m (pm 2 1)t. Let K 5 {x [ GF(pn)|x p 5 x}. We leave it as an easy exercise for the reader to show that K is a subfield of GF(pn). (Exercise 23). m Since the polynomial x p 2 x has at most pm zeros in GF(pn), we have |K| # pm. Let a 5 GF(pn)*. Then |at| 5 pm 2 1 and since (a t) p m21 5 1, it follows that at [ K. So, K is a subfield of GF(pn) of order pm. The uniqueness portion of the theorem follows from the observation that if GF(pn) had two distinct subfields of order pm, then the polynom mial x p 2 x would have more than pm zeros in GF(pn). This contradicts Corollary 3 of Theorem 16.2.

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Fields

Finally, suppose that F is a subfield of GF( pn). Then F is isomorphic to GF( pm) for some m and, by Theorem 21.5, n 5 [GF( pn):GF( p)] 5 [GF( pn):GF( pm)][GF( pm):GF( p)] 5 [GF( pn):GF( pm)]m. Thus, m divides n. Theorems 22.2 and 22.3, together with Theorem 4.3, make the task of finding the subfields of a finite field a simple exercise in arithmetic. EXAMPLE 3 Let F be the field of order 16 given in Example 1. Then there are exactly three subfields of F, and their orders are 2, 4, and 16. Obviously, the subfield of order 2 is {0, 1} and the subfield of order 16 is F itself. To find the subfield of order 4, we merely observe that the three nonzero elements of this subfield must be the cyclic subgroup of F* 5 x of order 3. So the subfield of order 4 is {0, 1, x5, x10} 5 {0, 1, x2 1 x, x2 1 x 1 1}. EXAMPLE 4 If F is a field of order 36 5 729 and a is a generator of F*, then the subfields of F are 1. GF(3) 5 {0} < a364 5 {0, 1, 2} 2. GF(9) 5 {0} < a91 3. GF(27) 5 {0} < a28 4. GF(729) 5 {0} < a. EXAMPLE 5 The subfield lattice of GF(224) is GF(224 ) GF(212 ) GF(28 ) GF(26 ) 4

GF(2 ) GF(2 3 ) 2

GF(2 )

GF(2)

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Exercises No pressure, no diamonds. MARY CASE

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

12. 13. 14. 15. 16.

17. 18. 19.

Find [GF(729):GF(9)] and [GF(64):GF(8)]. If m divides n, show that [GF( pn):GF( pm)] 5 n/m. Draw the lattice of subfields of GF(64). Let a be a zero of x3 1 x2 1 1 in some extension of Z2 . Find the multiplicative inverse of a 1 1 in Z2 3a4 . Let a be a zero of f(x) 5 x2 1 2x 1 1 in some extension of Z3. Find the other zero of f(x) in Z2 3a4 . Let a be a zero of f(x) 5 x3 1 x 1 1 in some extension of Z2 . Find the other zeros of f(x) in Z2 3a4 . Let K be a finite extension field of a finite field F. Show that there is an element a in K such that K 5 F(a). How many elements of the cyclic group GF(81)* are generators? Let f(x) be a cubic irreducible over Z2. Prove that the splitting field of f(x) over Z2 has order 8. Prove that the rings Z3[x]/x2 1 x 1 2 and Z3[x]/x2 1 2x 1 2 are isomorphic. Show that the Frobenius mapping f:GF( pn) S GF( pn), given by a S ap, is a ring automorphism of order n (that is, n is the smallest positive integer such that fn is the identity mapping). (This exercise is referred to in Chapter 32.) Determine the possible finite fields whose largest proper subfield is GF(25). Prove that the degree of any irreducible factor of x 8 2 x over Z2 is 1 or 3. Find the smallest field that has exactly 6 subfields. Show that x is a generator of the cyclic group (Z3[x]/x3 1 2x 1 1)*. Suppose that f(x) is a fifth-degree polynomial that is irreducible over Z2 . Prove that x is a generator of the cyclic group (Z2 3x4> f(x) ) *. Show that x is not a generator of the cyclic group (Z3[x]/x3 1 2x 1 2)*. Find one such generator. If f(x) is a cubic irreducible polynomial over Z3, prove that either x or 2x is a generator for the cyclic group (Z3[x]/ f(x))*. Prove the uniqueness portion of Theorem 22.3 using a grouptheoretic argument.

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Fields

20. Suppose that a and b belong to GF(81)*, with |a| 5 5 and |b| 5 16. Show that ab is a generator of GF(81)*. 21. Construct a field of order 9 and carry out the analysis as in Example 1, including the conversion table. 22. Show that any finite subgroup of the multiplicative group of a field is cyclic. 23. Show that the set K in the proof of Theorem 22.3 is a subfield. n 24. If g(x) is irreducible over GF( p) and g(x) divides x p 2 x, prove that deg g(x) divides n. 25. Use a purely group-theoretic argument to show that if F is a field n of order pn, then every element of F* is a zero of x p 2 x. (This exercise is referred to in the proof of Theorem 22.1.) 26. Draw the subfield lattices of GF(318) and of GF(230). 27. How does the subfield lattice of GF(230) compare with the subfield lattice of GF(330)? 28. If p(x) is a polynomial in Zp[x] with no multiple zeros, show that p(x) divides x pn 2 x for some n. 29. Suppose that p is a prime and p 2 2. Let a be a nonsquare in GF( p)—that is, a does not have the form b2 for any b in GF( p). Show that a is a nonsquare in GF( pn) if n is odd and that a is a square in GF( pn) if n is even. 30. Let f (x) be a cubic irreducible over Zp, where p is a prime. Prove that the splitting field of f(x) over Zp has order p3 or p6. 31. Show that every element of GF( pn) can be written in the form ap for some unique a in GF( pn). 32. Suppose that F is a field of order 1024 and F* 5 a. List the elements of each subfield of F. 33. Suppose that F is a field of order 125 and F* 5 a. Show that a62 5 21. 34. Show that no finite field is algebraically closed. n 35. Let E be the splitting field of f(x) 5 x p 2 x over Zp. Show that the set of zeros of f (x) in E is closed under addition, subtraction, multiplication, and division (by nonzero elements). (This exercise is referred to in the proof of Theorem 22.1.) 36. Suppose that L and K are subfields of GF( pn). If L has ps elements and K has pt elements, how many elements does L > K have? 37. Give an example to show that the mapping a S ap need not be an automorphism for arbitrary fields of prime characteristic p. 38. In the field GF(pn ) , show that for every positive divisor d of n, n xp 2 x has an irreducible factor over GF(p) of degree d.

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Computer Exercises Hardware: the parts of a computer that can be kicked. JEFF PESIS

Software for the computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software tests cubic polynomials over Zp for irreducibility. When a polynomial f(x) is irreducible, the software finds a generator for the cyclic group of nonzero elements of the field Zp[x]/ f(x) and creates a conversion table for addition and multiplication similar to Table 22.1. Run the program for p 5 2 and x3 1 x 1 1. Use the table to write (x2 1 x) (x2 1 x 1 1) (x2 1 1) as a power of x. Use the table to write x12 in additive form. 2. This software tests fourth-degree polynomials over Zp for p 5 2 or 3 for irreducibility. When a polynomial f(x) is irreducible, the software finds a generator for the cyclic group of nonzero elements of the field Zp 3x4>  f(x) and creates a conversion table for addition and multiplication similar to Table 22.1 in the text. Run the program for p 5 2 and x4 1 x 1 1. Use the table to write (x2 1 x) (x2 1 x 1 1) (x3 1 1) as a power of x. Use the table to write x23 in additive form.

Suggested Reading Judy L. Smith and J. A. Gallian, “Factoring Finite Factor Rings,” Mathematics Magazine 58 (1985): 93–95. This paper gives an algorithm for finding the group of units of the ring F[x]/g(x)m.

L .E. Dickson One of the books [written by L. E. Dickson] is his major, three-volume History of the Theory of Numbers which would be a life’s work by itself for a more ordinary man. A. A. ALBERT, Bulletin of American Mathematical Society

LEONARD EUGENE DICKSON was born in Independence, Iowa, on January 22, 1874. In 1896, he received the first Ph.D. to be awarded in mathematics at the University of Chicago. After spending a few years at the University of California and the University of Texas, he was appointed to the faculty at Chicago and remained there until his retirement in 1939. Dickson was one of the most prolific mathematicians of the 20th century, writing 267 research papers and 18 books. His principal interests were matrix groups, finite fields, algebra, and number theory. Dickson had a disdainful attitude toward applicable mathematics; he would often say, “Thank God that number theory is unsullied by any applications.” He also had a sense of

392

humor. Dickson would often mention his honeymoon: “It was a great success,” he said, “except that I only got two research papers written.” Dickson received many honors in his career. He was the first to be awarded the prize from the American Association for the Advancement of Science for the most notable contribution to the advancement of science, and the first to receive the Cole Prize in algebra from the American Mathematical Society. The University of Chicago has research instructorships named after him. Dickson died on January 17, 1954. For more information about Dickson, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Geometric 23 Constructions At the age of eleven, I began Euclid. . . . This was one of the great events of my life, as dazzling as first love. BERTRAND RUSSELL

Historical Discussion of Geometric Constructions The ancient Greeks were fond of geometric constructions. They were especially interested in constructions that could be achieved using only a straightedge without markings and a compass. They knew, for example, that any angle can be bisected, and they knew how to construct an equilateral triangle, a square, a regular pentagon, and a regular hexagon. But they did not know how to trisect every angle or how to construct a regular seven-sided polygon (heptagon). Another problem that they attempted was the duplication of the cube—that is, given any cube, they tried to construct a new cube having twice the volume of the given one using only an unmarked straightedge and a compass. Legend has it that the ancient Athenians were told by the oracle at Delos that a plague would end if they constructed a new altar to Apollo in the shape of a cube with double the volume of the old altar, which was also a cube. Besides “doubling the cube,” the Greeks also attempted to “square the circle”—to construct a square with area equal to that of a given circle. They knew how to solve all these problems using other means, such as a compass and a straightedge with two marks, or an unmarked straightedge and a spiral, but they could not achieve any of the constructions with a compass and an unmarked straightedge alone. These problems vexed mathematicians for over 2000 years. The resolution of these perplexities was made possible when they were transferred from questions of geometry to questions of algebra in the 19th century. The first of the famous problems of antiquity to be solved was that of the construction of regular polygons. It had been known since Euclid that regular polygons with a number of sides of the form 2k, 2k ? 3, 2k ? 5, and 2k ? 3 ? 5 could be constructed, and it was believed that no others were 393

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possible. In 1796, while still a teenager, Gauss proved that the 17-sided regular polygon is constructible. In 1801, Gauss asserted that a regular polygon of n sides is constructible if and only if n hass the form 2kp1 p2 ? ? ? pi, where the p’s are distinct primes of the form 22 1 1. We provide a proof of this statement in Theorem 33.5. Thus, regular polygons with 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, and 20 sides are possible to construct, whereas those with 7, 9, 11, 13, 14, 18, and 19 sides are not. How these constructions can be effected is another matter. One person spent 10 years trying to determine a way to construct the 65,537-sided polygon. Gauss’s result on the constructibility of regular n-gons eliminated another of the famous unsolved problems because the ability to trisect a 60° angle enables one to construct a regular 9-gon. Thus, there is no method for trisecting a 60° angle with an unmarked straightedge and a compass. In 1837, Wantzel proved that it was not possible to double the cube. The problem of the squaring of a circle resisted all attempts until 1882, when Ferdinand Lindemann proved that p is transcendental since, as we will show, all constructible numbers are algebraic.

Constructible Numbers With the field theory we now have, it is an easy matter to solve the following problem: Given an unmarked straightedge, a compass, and a unit length, what other lengths can be constructed? To begin, we call a real number a constructible if, by means of an unmarked straightedge, a compass, and a line segment of length 1, we can construct a line segment of length |a| in a finite number of steps. It follows from plane geometry that if a and b (b 2 0) are constructible numbers, then so are a 1 b, a 2 b, a ? b, and a/b. (See the exercises for hints.) Thus, the set of constructible numbers contains Q and is a subfield of the real numbers. What we desire is an algebraic characterization of this field. To derive such a characterization, let F be any subfield of the reals. Call the subset {(x, y)[ R2| x, y [ F} of the real plane the plane of F, call any line joining two points in the plane of F a line in F, and call any circle whose center is in the plane of F and whose radius is in F a circle in F. Then a line in F has an equation of the form ax 1 by 1 c 5 0,

where a, b, c [ F,

and a circle in F has an equation of the form x2 1 y2 1 ax 1 by 1 c 5 0,

where a, b, c [ F.

In particular, note that to find the point of intersection of a pair of lines in F or the points of intersection of a line in F and a circle in F, one

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need only solve a linear or quadratic equation in F. We now come to the crucial question. Starting with points in the plane of some field F, which points in the real plane can be obtained with an unmarked straightedge and a compass? Well, there are only three ways to construct points, starting with points in the plane of F. 1. Intersect two lines in F. 2. Intersect a circle in F and a line in F. 3. Intersect two circles in F. In case 1, we do not obtain any new points, because two lines in F intersect in a point in the plane of F. In case 2, the point of intersection is the solution to either a linear equation in F or a quadratic equation in F. So, the point lies in the plane of F or in the plane of F("a ), where a [ F and a is positive. In case 3, no new points are obtained, because, if the two circles are given by x2 1 y2 1 ax 1 by 1 c 5 0 and x2 1 y2 1 a9x 1 b9y 1 c9 5 0, then we have (a 2 a9)x 1 (b 2 b9)y 1 (c 2 c9) 5 0, which is a line in F. So, the points of intersection are in F. It follows, then, that the only points in the real plane that can be constructed from the plane of a field F are those whose coordinates lie in fields of the form F("a), where a [ F and a is positive. Of course, we can start over with F1 5 F("a ) and construct points whose coordinates lie in fields of the form F2 5 F1("b ), where b [ F1 and b is positive. Continuing in this fashion, we see that a real number c is constructible if and only if there is a series of fields Q 5 F1 # F2 # ? ? ? # Fn # R such that Fi11 5 Fi ("ai), where ai [ Fi and c [ Fn. Since [Fi11:Fi] 5 1 or 2, we see by Theorem 21.5 that if c is constructible, then [Q(c):Q] 5 2k for some nonnegative integer k. We now dispatch the problems that plagued the Greeks. Consider doubling the cube of volume 1. The enlarged cube would have an edge of 3 3 length " 2. But [Q(" 2):Q] 5 3, so such a cube cannot be constructed. Next consider the possibility of trisecting a 60° angle. If it were possible to trisect an angle of 60°, then cos 20° would be constructible. (See Figure 23.1.) In particular, [Q(cos 20°):Q] 5 2k for some k. Now, using the trigonometric identity cos 3u 5 4 cos3 u 2 3 cos u, with u 5 20°, we

(cos 20°, sin 20°)

(0, 0)

(1, 0)

Figure 23.1

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Fields

see that 1/2 5 4 cos3 20° 2 3 cos 20°, so that cos 20° is a zero of 8x3 2 6x 2 1. But, since 8x3 2 6x 2 1 is irreducible over Q (see Exercise 13), we must also have [Q(cos 20°):Q] 5 3. This contradiction shows that trisecting a 60° angle is impossible. The remaining problems are relegated to the reader as Exercises 14, 15, and 17.

Angle-Trisectors and Circle-Squarers Down through the centuries, hundreds of people have claimed to have achieved one or more of the impossible constructions. In 1775, the Paris Academy, so overwhelmed with these claims, passed a resolution to no longer examine these claims or claims of machines purported to exhibit perpetual motion. Although it has been more than 100 years since the last of the constructions was shown to be impossible, there continues to be a steady parade of people who claim to have done one or more of them. Most of these people have heard that this is impossible but have refused to believe it. One person insisted that he could trisect any angle with a straightedge alone [2, p. 158]. Another found his trisection in 1973 after 12,000 hours of work [2, p. 80]. One got his from God [2, p. 73]. In 1971, a person with a Ph.D. in mathematics asserted that he had a valid trisection method [2, p. 127]. Many people have claimed the hat trick: trisecting the angle, doubling the cube, and squaring the circle. Two men who did this in 1961 succeeded in having their accomplishment noted in the Congressional Record [2, p. 110]. Occasionally, newspapers and magazines have run stories about “doing the impossible,” often giving the impression that the construction may be valid. Many angle-trisectors and circle-squarers have had their work published at their own expense and distributed to colleges and universities. One had his printed in four languages! There are two delightful books written by mathematicians about their encounters with these people. The books are full of wit, charm, and humor ([1] and [2]).

Exercises Only prove to me that it is impossible, and I will set about it this very evening. Spoken by a member of the audience after De Morgan gave a lecture on the impossibility of squaring the circle.

1. If a and b are constructible numbers and a $ b . 0, give a geometric proof that a 1 b and a 2 b are constructible.

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2. If a and b are constructible, give a geometric proof that ab is constructible. (Hint: Consider the following figure. Notice that all segments in the figure can be made with an unmarked straightedge and a compass.)

a

1 b

3. Prove that if c is a constructible number, then so is "|c|. (Hint: Consider the following semicircle with diameter 1 1 |c|.) (This exercise is referred to in Chapter 33.) α d

α

⏐c⏐

1

4. If a and b (b 2 0) are constructible numbers, give a geometric proof that a/b is constructible. (Hint: Consider the following figure.) b

0

1 a

5. Prove that sin u is constructible if and only if cos u is constructible. 6. Prove that an angle u is constructible if and only if sin u is constructible. 7. Prove that cos 2u is constructible if and only if cos u is constructible. 8. Prove that 30° is a constructible angle. 9. Prove that a 45° angle can be trisected with an unmarked straightedge and a compass. 10. Prove that a 40° angle is not constructible. 11. Show that the point of intersection of two lines in the plane of a field F lies in the plane of F.

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12. Show that the points of intersection of a circle in the plane of a field F and a line in the plane of F are points in the plane of F or in the plane of F("a ), where a [ F and a is positive. Give an example of a circle and a line in the plane of Q whose points of intersection are not in the plane of Q. 13. Prove that 8x3 2 6x 2 1 is irreducible over Q. 14. Use the fact that 8 cos3(2p/7) 1 4 cos2(2p/7) 2 4 cos(2p/7) 2 1 5 0 to prove that a regular seven-sided polygon is not constructible with an unmarked straightedge and a compass. 15. Show that a regular 9-gon cannot be constructed with an unmarked straightedge and a compass. 16. Show that if a regular n-gon is constructible, then so is a regular 2n-gon. 17. (Squaring the Circle) Show that it is impossible to construct, with an unmarked straightedge and a compass, a square whose area equals that of a circle of radius 1. You may use the fact that p is transcendental over Q. 18. Use the fact that 4 cos2(2p/5) 1 2 cos(2p/5) 2 1 5 0 to prove that a regular pentagon is constructible. 19. Can the cube be “tripled”? 20. Can the cube be “quadrupled”? 21. Can the circle be “cubed”? 22. If a, b, and c are constructible, show that the real roots of ax2 1 bx 1 c are constructible.

References 1. Augustus De Morgan, A Budget of Paradoxes, 2nd ed., Salem, N.H.: Ayer, 1915. 2. Underwood Dudley, A Budget of Trisections, New York: Springer-Verlag, 1987.

Suggested Website http://en.wikipedia.org/wiki/Squaring_the_circle This website provides an excellent account of efforts to square the circle, and links for articles about trisecting the angle and doubling the cube.

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Supplementary Exercises for Chapters 19–23 Difficulties strengthen the mind, as labor does the body. SENECA

True/false questions for Chapters 19–23 are available on the Web at http://www.d.umn.edu/~jgallian/TF 2 1 has no multiple zeros in any extension of Z3. 1. Show that 2. Suppose that p(x) is a quadratic polynomial with rational coefficients and is irreducible over Q. Show that p(x) has two zeros in Q[x]/p(x). 3. Let F be a finite field of order q and let a be a nonzero element in F. If n divides q 2 1, prove that the equation x n 5 a has either no solutions in F or n distinct solutions in F. 4. Without using the Primitive Element Theorem, prove that if [K:F] is prime, then K has a primitive element. 5. Let a be a zero of x2 1 x 1 1. Express (5a2 1 2)/a in the form c 1 ba, where c and b are rational. 4 6. Describe the elements of the extension Q(" 2) over the field Q("2). 7. If [F(a):F] 5 5, find [F(a3):F]. Does your argument apply equally well if a3 is replaced with a2 and a4? 8. If p(x) [ F[x] and deg p(x) 5 n, show that the splitting field for p(x) over F has degree at most n!. 9. Let a be a nonzero algebraic element over F of degree n. Show that a21 is also algebraic over F of degree n. 10. Prove that p2 2 1 is algebraic over Q(p3). 11. If ab is algebraic over F and b 2 0, prove that a is algebraic over F(b). 12. Let E be an algebraic extension of a field F. If R is a ring and E $ R $ F, show that R must be a field. 13. If a is transcendental over F, show that every element of F(a) that is not in F is transcendental over F. 14. What is the order of the splitting field of x5 1 x4 1 1 5 (x2 1 x 1 1) ? (x3 1 x 1 1) over Z2? 15. Show that a finite extension of a finite field is a simple extension. 16. Let R be an integral domain that contains a field F as a subring. If R is finite-dimensional when viewed as a vector space over F, prove that R is a field. x50

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17. Show that it is impossible to find a basis for the vector space of n 3 n (n . 1) matrices such that each pair of elements in the basis commutes under multiplication. 18. Let Pn 5 {anxn 1 an21xn21 1 ? ? ? 1 a1x 1 a0 | each ai is a real number}. Is it possible to have a basis for Pn such that every element of the basis has x as a factor? 19. Find a basis for the vector space { f [ P3|f (0) 5 0}. (See Exercise 18 for notation.) 20. Given that f is a polynomial of degree n in Pn, show that { f, f 9, f 0, . . . , f (n)} is a basis for Pn. ( f (k) denotes the kth derivative of f.) 21. Suppose that K is a field extension of a field F of characteristic n p 2 0. Let L 5 {a [ K|a p [ F for some nonnegative integer n}. Prove that L is a subfield of K that contains F. 22. In which fields does x n 2 x have a multiple zero?

PA R T

5

Special Topics

For online student resources, visit this textbook’s website at http://college.hmco.com/PIC/gallian7e

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24 Sylow Theorems

Generally these three results are implied by the expression “Sylow’s Theorem.” All of them are of fundamental importance. In fact, if the theorems of group theory were arranged in order of their importance Sylow’s Theorem might reasonably occupy the second place—coming next to Lagrange’s Theorem in such an arrangement. G. A. MILLER, Theory and Application

of Finite Groups

Conjugacy Classes In this chapter, we derive several important arithmetic relationships between a group and certain of its subgroups. Recall from Chapter 7 that Lagrange’s Theorem was proved by showing that cosets of a subgroup partition the group. Another fruitful method of partitioning the elements of a group is by way of conjugacy classes. Definition Conjugacy Class of a

Let a and b be elements of a group G. We say that a and b are conjugate in G (and call b a conjugate of a) if xax21 5 b for some x in G. The conjugacy class of a is the set cl(a) 5 {xax21 | x [ G}.

We leave it to the reader (Exercise 1) to prove that conjugacy is an equivalence relation on G, and that the conjugacy class of a is the equivalence class of a under conjugacy. Thus, we may partition any group into disjoint conjugacy classes. Let’s look at one example. In D4 we have cl(H) 5 {R0HR021, R90 HR9021, R180 HR18021, R270HR27021, HHH21, VHV21, DHD21, D9HD921} 5 {H, V}. Similarly, one may verify that cl(R0) 5 {R0}, cl(R90) 5 {R90, R270} 5 cl(R270), cl(R180) 5 {R180}, cl(V) 5 {V, H} 5 cl(H), cl(D) 5 {D, D9} 5 cl(D9). 403

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Theorem 24.1 gives an arithmetic relationship between the size of the conjugacy class of a and the size of C(a), the centralizer of a. Theorem 24.1 The Number of Conjugates of a Let G be a finite group and let a be an element of G. Then, |cl(a)| 5 |G:C(a)|.

PROOF Consider the function T that sends the coset xC(a) to the conjugate xax21 of a. A routine calculation shows that T is well defined, is one-to-one, and maps the set of left cosets onto the conjugacy class of a. Thus, the number of conjugates of a is the index of the centralizer of a. Corollary 1 |cl(a)| Divides |G| In a finite group, |cl(a)| divides |G|.

The Class Equation Since the conjugacy classes partition a group, the following important counting principle is a corollary to Theorem 24.1. Corollary 2 The Class Equation For any finite group G,

|G| 5 S |G:C(a)|, where the sum runs over one element a from each conjugacy class of G.

In finite group theory, counting principles such as this corollary are powerful tools.† Theorem 24.2 is the single most important fact about finite groups of prime-power order (a group of order pn, where p is a prime, is called a p-group).

†“Never

underestimate a theorem that counts something.” John Fraleigh, A First Course in Abstract Algebra.

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Theorem 24.2 p-Groups Have Nontrivial Centers Let G be a nontrivial finite group whose order is a power of a prime p. Then Z(G) has more than one element.

PROOF First observe that cl(a) 5 {a} if and only if a [ Z(G) (see Exercise 4). Thus, by culling out these elements, we may write the class equation in the form |G| 5 |Z(G)| 1 S|G:C(a)|, where the sum runs over representatives of all conjugacy classes with more than one element (this set may be empty). But |G:C(a)| 5 |G|/|C(a)|, so each term in S|G:C(a)| has the form pk with k $ 1. Hence, |G| 2 S|G:C(a)| 5 |Z(G)|, where each term on the left is divisible by p. It follows, then, that p also divides |Z(G)|, and hence |Z(G)| 2 1. Corollary Groups of Order p2 Are Abelian If |G| 5 p2, where p is prime, then G is Abelian.

PROOF By Theorem 24.2 and Lagrange’s Theorem, |Z(G)| 5 p or p2. If |Z(G)| 5 p2, then G 5 Z(G) and G is Abelian. If |Z(G)| 5 p, then |G/Z(G)| 5 p, so that G/Z(G) is cyclic. But, then, by Theorem 9.3, G is Abelian.

The Probability That Two Elements Commute Before proceeding to the main goal of this chapter, we pause for an interesting application of Theorem 24.1 and the class equation. (Our discussion is based on [1] and [2].) Suppose we select two elements at random (with replacement) from a finite group. What is the probability that these two elements commute? Well, suppose that G is a finite group of order n. Then the probability Pr(G) that two elements selected at random from G commute is |K|/n2, where K 5 {(x, y) [ G % G | xy 5 yx}. Now notice that for each x [ G we have (x, y) [ K if and only if y [ C(x). Thus, 0 K 0 5 a 0 C(x) 0. x[G

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Also, it follows from Theorem 24.1 that if x and y are in the same conjugacy class, then |C(x)| 5 |C(y)| (see Exercise 53). If, for example, cl(a) 5 {a1, a2, . . . , at}, then |C(a1)| 1 |C(a2)| 1 ? ? ? 1 |C(at)| 5 t|C(a)| 5 |G:C(a)| |C(a)| 5 |G| 5 n. So, by choosing one representative from each conjugacy class, say, x1, x2, . . . , xm, we have m

0 K 0 5 a 0 C(x) 0 5 a 0 G:C(xi ) 0 0 C(xi ) 0 5 m # n. i51 xPG Thus, the answer to our question is mn/n2 5 m/n, where m is the number of conjugacy classes in G and n is the number of elements of G. Obviously, when G is non-Abelian, Pr(G) is less than 1. But how much less than 1? Clearly, the more conjugacy classes there are, the larger Pr(G) is. Consequently, Pr(G) is large when the sizes of the conjugacy classes are small. Noting that |cl(a)| 5 1 if and only if a [ Z(G), we obtain the maximum number of conjugacy classes when |Z(G)| is as large as possible and all other conjugacy classes have exactly two elements in each. Since G is non-Abelian, it follows from Theorem 9.3 that |G/Z(G)| $ 4 and, therefore, |Z(G)| # |G|/4. Thus, in the extreme case, we would have |Z(G)| 5 |G|/4, and the remaining (3/4)|G| elements would be distributed in conjugacy classes with two elements each. So, in a non-Abelian group, the number of conjugacy classes is no more than |G|/4 1 (1/2)(3/4)|G|, and Pr(G) is less than or equal to 5/8. The dihedral group D4 is an example of a group that has probability equal to 5/8.

The Sylow Theorems Now to the Sylow theorems. Recall that the converse of Lagrange’s Theorem is false; that is, if G is a group of order m and n divides m, G need not have a subgroup of order n. Our next theorem is a partial converse of Lagrange’s Theorem. It, as well as Theorem 24.2, was first proved by the Norwegian mathematician Ludwig Sylow (1832–1918). Sylow’s Theorem and Lagrange’s Theorem are the two most important results in finite group theory. The first gives a sufficient condition for the existence of subgroups, and the second gives a necessary condition. Theorem 24.3 Existence of Subgroups of Prime-Power Order (Sylow’s First Theorem, 1872) Let G be a finite group and let p be a prime. If p k divides |G|, then G has at least one subgroup of order p k.

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PROOF We proceed by induction on |G|. If |G| 5 1, Theorem 24.3 is trivially true. Now assume that the statement is true for all groups of order less than |G|. If G has a proper subgroup H such that pk divides |H|, then, by our inductive assumption, H has a subgroup of order pk and we are done. Thus, we may henceforth assume that pk does not divide the order of any proper subgroup of G. Next, consider the class equation for G in the form |G| 5 |Z(G)| 1 S|G:C(a)|, where we sum over a representative of each conjugacy class cl(a), where a o Z(G). Since pk divides |G| 5 |G:C(a)| |C(a)| and pk does not divide |C(a)|, we know that p must divide |G:C(a)| for all a o Z(G). It then follows from the class equation that p divides |Z(G)|. The Fundamental Theorem of Finite Abelian Groups (Theorem 11.1), or Theorem 9.5, then guarantees that Z(G) contains an element of order p, say x. Since x is in the center of G, x is a normal subgroup of G, and we may form the factor group G/x. Now observe that pk21 divides |G/x|. Thus, by the induction hypothesis, G/x has a subgroup of order pk21 and, by Exercise 49 in Chapter 10, this subgroup has the form H/x, where H is a subgroup of G. Finally, note that |H/x| 5 pk21 and |x| 5 p imply that |H| 5 pk, and this completes the proof. Let’s be sure we understand exactly what Sylow’s First Theorem means. Say we have a group G of order 23 ? 32 ? 54 ? 7. Then Sylow’s First Theorem says that G must have at least one subgroup of each of the following orders: 2, 4, 8, 3, 9, 5, 25, 125, 625, and 7. On the other hand, Sylow’s First Theorem tells us nothing about the possible existence of subgroups of order 6, 10, 15, 30, or any other divisor of |G| that has two or more distinct prime factors. Because certain subgroups guaranteed by Sylow’s First Theorem play a central role in the theory of finite groups, they are given a special name. Definition Sylow p-Subgroup

Let G be a finite group and let p be a prime divisor of |G|. If pk divides |G| and pk11 does not divide |G|, then any subgroup of G of order pk is called a Sylow p-subgroup of G.

So, returning to our group G of order 23 ? 32 ? 54 ? 7, we call any subgroup of order 8 a Sylow 2-subgroup of G, any subgroup of order 625 a Sylow 5-subgroup of G, and so on. Notice that a Sylow p-subgroup of G is a subgroup whose order is the largest power of p consistent with Lagrange’s Theorem.

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Since any subgroup of order p is cyclic, we have the following generalization of Theorem 9.5, first proved by Cauchy in 1845. His proof ran nine pages! Corollary Cauchy’s Theorem Let G be a finite group and let p be a prime that divides the order of G. Then G has an element of order p.

Sylow’s First Theorem is so fundamental to finite group theory that many different proofs of it have been published over the years [our proof is essentially the one given by Georg Frobenius (1849–1917) in 1895]. Likewise, there are scores of generalizations of Sylow’s Theorem. Observe that the corollary to the Fundamental Theorem of Finite Abelian Groups and Sylow’s First Theorem show that the converse of Lagrange’s Theorem is true for all finite Abelian groups and all finite groups of prime-power order. There are two more Sylow theorems that are extremely valuable tools in finite group theory. But first we introduce a new term. Definition Conjugate Subgroups

Let H and K be subgroups of a group G. We say that H and K are conjugate in G if there is an element g in G such that H 5 gKg21.

Recall from Chapter 7 that if G is a finite group of permutations on a set S and i [ S, then orbG(i) 5 {f(i) | f [ G} and |orbG(i)| divides |G|. Theorem 24.4 Sylow’s Second Theorem If H is a subgroup of a finite group G and |H| is a power of a prime p, then H is contained in some Sylow p-subgroup of G.

PROOF Let K be a Sylow p-subgroup of G and let C 5 {K1, K2, . . . , Kn} with K 5 K1 be the set of all conjugates of K in G. Since conjugation is an automorphism, each element of C is a Sylow p-subgroup of G. Let SC denote the group of all permutations of C. For each g [ G, define fg:C S C by fg(Ki) 5 gKig21. It is easy to show that each fg [ SC. Now define a mapping T:G S SC by T(g) 5 fg. Since fgh(Ki) 5 (gh)K i (gh) 21 5 g(hK i h 21 )g 21 5 gf h (K i )g 21 5 f g (f h (K i )) 5 (f g f h )(Ki), we have fgh 5 fgfh, and therefore T is a homomorphism from G to SC.

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Next consider T(H), the image of H under T. Since |H| is a power of p, so is |T(H)| (see property 6 of Theorem 10.2). Thus, by the OrbitStabilizer Theorem (Theorem 7.3), for each i, |orbT(H)(Ki)| divides |T(H)|, so that |orbT(H)(Ki)| is a power of p. Now we ask: Under what condition does |orbT(H)(Ki)| 5 1? Well, |orbT(H)(Ki)| 5 1 means that fg(Ki) 5 gKig21 5 Ki for all g [ H; that is, |orbT(H)(Ki)| 5 1 if and only if H # N(Ki). But the only elements of N(Ki) that have orders that are powers of p are those of Ki (see Exercise 9). Thus, |orbT(H)(Ki)| 5 1 if and only if H # Ki. So, to complete the proof, all we need to do is show that for some i, |orbT(H)(Ki)| 5 1. Analogous to Theorem 24.1, we have |C| 5 |G:N(K)| (see Exercise 21). And since |G:K| 5 |G:N(K)||N(K):K| is not divisible by p, neither is |C|. Because the orbits partition C, |C| is the sum of powers of p. If no orbit has size 1, then p divides each summand and, therefore, p divides |C|, which is a contradiction. Thus, there is an orbit of size 1, and the proof is complete. Theorem 24.5 Sylow’s Third Theorem Let p be a prime and let G be a group of order pkm, where p does not divide m. Then the number n of Sylow p-subgroups of G is equal to 1 modulo p and divides m. Furthermore, any two Sylow p-subgroups of G are conjugate.

PROOF Let K be any Sylow p-subgroup of G and let C 5 {K 1 , K 2 , . . . , Kn} with K 5 K1 be the set of all conjugates of K in G. We first prove that n mod p 5 1. Let SC and T be as in the proof of Theorem 24.4. This time we consider T(K), the image of K under T. As before, we have |orbT(K)(Ki)| is a power of p for each i and |orbT(K)(Ki)| 5 1 if and only if K # Ki. Thus, |orbT(K)(K1)| 5 1 and |orbT(K)(Ki)| is a power of p greater than 1 for all i 2 1. Since the orbits partition C, it follows that, modulo p, n 5 |C| 5 1. Next we show that every Sylow p-subgroup of G belongs to C. To do this, suppose that H is a Sylow p-subgroup of G that is not in C. Let SC and T be as in the proof of Theorem 24.4, and this time consider T(H). As in the previous paragraph, |C| is the sum of the orbits’ sizes under the action of T(H). However, no orbit has size 1, since H is not in C. Thus, |C| is a sum of terms each divisible by p, so that, modulo p, n 5 |C| 5 0. This contradiction proves that H belongs to C, and that n is the number of Sylow p-subgroups of G. Finally, that n divides |G| follows directly from the fact that n 5 |G:N(K)| (see Exercise 21).

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It is convenient to let np denote the number of Sylow p-subgroups of a group. Observe that the first portion of Sylow’s Third Theorem is a counting principle.† As an important consequence of Sylow’s Third Theorem, we have the following corollary. Corollary A Unique Sylow p-Subgroup Is Normal A Sylow p-subgroup of a finite group G is a normal subgroup of G if and only if it is the only Sylow p-subgroup of G.

We illustrate Sylow’s Third Theorem with two examples. EXAMPLE 1 Consider the Sylow 2-subgroups of S3. They are {(1), (12)}, {(1), (23)}, and {(1), (13)}. According to Sylow’s Third Theorem, we should be able to obtain the latter two of these from the first by conjugation. Indeed, (13){(1), (12)}(13)21 5 {(1), (23)}, (23){(1), (12)}(23)21 5 {(1), (13)}. EXAMPLE 2 Consider the Sylow 3-subgroups of A4. They are {a1, a5, a9}, {a1, a6, a11}, {a1, a7, a12}, and {a1, a8, a10}. (See the table on page 107.) Then, a2{a1, a5, a9}a221 5 {a1, a7, a12}, a3{a1, a5, a9}a321 5 {a1, a8, a10}, a4{a1, a5, a9}a421 5 {a1, a6, a11}. Thus, the number of Sylow 3-subgroups is 1 modulo 3, and the four Sylow 3-subgroups are conjugate. Figure 24.1 shows the subgroup lattices for S3 and A4. We have connected the Sylow p-groups with dashed circles to indicate that they belong to one orbit under conjugation. Notice that the three subgroups of order 2 in A4 are contained in a Sylow 2-group, as required by Sylow’s Second Theorem. As it happens, these three subgroups also belong to one orbit under conjugation, but this is not a consequence of Sylow’s Third Theorem. In contrast to the two preceding examples, observe that the dihedral group of order 12 has seven subgroups of order 2, but that conjugating {R0, R180} does not yield any of the other six. (Why?) †“Whenever

Mathematics.

you can, count.” Sir Francis Galton (1822–1911), The World of

24 | Sylow Theorems

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S3

A3 = <(123)>

<(12)>

<(13)> <(23)>

<(1)> A4 <α8 >

{α 1, α 2, α 3, α 4} < α5 > < α 2>

< α 4>

< α 7> < α6 >

< α3 >

< α 1>

Figure 24.1 Lattices of subgroups for S3 and A4.

Applications of Sylow Theorems A few numerical examples will make the Sylow theorems come to life. Say G is a group of order 40. What do the Sylow theorems tell us about G? A great deal! Since 1 is the only divisor of 40 that is congruent to 1 modulo 5, we know that G has exactly one subgroup of order 5, and therefore it is normal. Similarly, G has either one or five subgroups of order 8. If there is only one subgroup of order 8, it is normal. If there are five subgroups of order 8, none is normal and all five can be obtained by starting with any particular one, say H, and computing xHx21 for various x’s. Finally, if we let K denote the normal subgroup of order 5 and let H denote any subgroup of order 8, then G 5 HK. (See Exercise 7, Supplementary Exercises for Chapters 5–8.) If H happens to be normal, we can say even more: G 5 H 3 K. What about a group G of order 30? It must have either one or six subgroups of order 5 and one or ten subgroups of order 3. However, G cannot have both six subgroups of order 5 and 10 subgroups of order 3 (for then G would have more than 30 elements). Thus, G has one subgroup of order 3 and one of order 5, and at least one of these is normal

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in G. It follows, then, that the product of a subgroup of order 3 and one of order 5 is a group of order 15 that is both cyclic (Exercise 25) and normal (Exercise 7 in Chapter 9) in G. [This, in turn, implies that both the subgroup of order 3 and the subgroup of order 5 are normal in G (Exercise 57 in Chapter 9).] So, if we let y be a generator of the cyclic subgroup of order 15 and let x be an element of order 2 (the existence of which is guaranteed by Cauchy’s Theorem), we see that G 5 {x iy j | 0 # i # 1, 0 # j # 14}. Note that in these two examples we were able to deduce all of this information from knowing only the order of the group—so many conclusions from one assumption! This is the beauty of finite group theory. In Chapter 7 we saw that the only group (up to isomorphism) of prime order p is Zp. As a further illustration of the power of the Sylow theorems, we next give a sufficient condition that guarantees that a group of order pq, where p and q are primes, must be Zpq. Theorem 24.6 Cyclic Groups of Order pq If G is a group of order pq, where p and q are primes, p , q, and p does not divide q 2 1, then G is cyclic. In particular, G is isomorphic to Zpq.

PROOF Let H be a Sylow p-subgroup of G and let K be a Sylow q-subgroup of G. Sylow’s Third Theorem states that the number of Sylow p-subgroups of G is of the form 1 1 kp and divides pq. So 1 1 kp is equal to 1, p, q, or pq. From this and the fact that p B q 2 1, it follows that k 5 0, and therefore H is the only Sylow p-subgroup of G. Similarly, there is only one Sylow q-subgroup of G. Thus, by the corollary to Theorem 24.5, H and K are normal subgroups of G. Let H 5 x and K 5 y. To show that G is cyclic, it suffices to show that x and y commute, for then |xy| 5 |x||y| 5 pq. But observe that, since H and K are normal, we have xyx21y21 5 (xyx21)y21 [ Ky21 5 K and xyx21y21 5 x(yx21y21) [ xH 5 H. Thus, xyx21y21 [ K > H 5 {e}, and hence xy 5 yx. Theorem 24.6 demonstrates the power of the Sylow theorems in classifying the finite groups whose orders have small numbers of prime factors. Results along the lines of Theorem 24.6 exist for groups of orders p2q, p2q2, p3, and p4, where p and q are prime.

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16

Figure 24.2 The number of groups of a given order up to 100.

For your amusement, Figure 24.2 lists the number of nonisomorphic groups with orders at most 100. Note in particular the large number of groups of order 64. Also observe that, generally speaking, it is not the size of the group that gives rise to a large number of groups of that size but the number of prime factors involved. In all, there are 1047 nonisomorphic groups with 100 or fewer elements. Contrast this with the fact reported in 1989 that there are 2328 groups of order 128 and 56,092 of order 256 [3]. The number of groups of any order up to 1000, except 512, is given at http://people.csse.uwa.edu.au/gordon/remote/cubcay/. Estimates put the number of groups of order 512 at more than one million. As a final application of the Sylow theorems, you might enjoy seeing a determination of the groups of order 99, 66, and 255. In fact, our arguments serve as a good review of much of our work in group theory. EXAMPLE 3 Determination of the Groups of Order 99 Suppose that G is a group of order 99. Let H be a Sylow 3-subgroup of G and let K be a Sylow 11-subgroup of G. Since 1 is the only positive divisor of 99 that is equal to 1 modulo 11, we know from Sylow’s Third Theorem and its corollary that K is normal in G. Similarly, H is normal in G. It follows, by the argument used in the proof of Theorem 24.6, that elements from H and K commute, and therefore G 5 H 3 K. Since both H and K are Abelian, G is also Abelian. Thus, G is isomorphic to Z99 or Z3 % Z33. EXAMPLE 4 Determination of the Groups of Order 66 Suppose that G is a group of order 66. Let H be a Sylow 3-subgroup of G and let K be a Sylow 11-subgroup of G. Since 1 is the only positive divisor of 66 that is equal to 1 modulo 11, we know that K is normal in G. Thus, HK is a subgroup of G of order 33 (Exercise 55 in Chapter 9 and Exercise 7, Supplementary Exercises for Chapters 5–8). Since any

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group of order 33 is cyclic (Theorem 24.6), we may write HK 5 x. Next, let y [ G and |y| 5 2. Since x has index 2 in G, we know it is normal. So yxy21 5 xi for some i from 1 to 32. Then, yx 5 xiy and, since every member of G is of the form xsyt, the structure of G is completely determined by the value of i. We claim that there are only four possibilities for i. To prove this, observe that |xi| 5 |x| (Exercise 5, Supplementary Exercises for Chapters 1–4). Thus, i and 33 are relatively prime. But also, since y has order 2, 2

x 5 y21( yxy21)y 5 y21xiy 5 yxiy21 5 (yxy21)i 5 (xi)i 5 x i . 2

So xi 21 5 e and therefore 33 divides i2 2 1. From this it follows that 11 divides i 6 1, and therefore i 5 0 6 1, i 5 11 6 1, i 5 22 6 1, or i 5 33 6 1. Putting this together with the other information we have about i, we see that i 5 1, 10, 23, or 32. This proves that there are at most four groups of order 66. To prove that there are exactly four such groups, we simply observe that Z66, D33, D11 % Z3, and D3 % Z11 each has order 66 and that no two are isomorphic. For example, D11 % Z3 has 11 elements of order 2, whereas D3 % Z11 has only three elements of order 2. (See Exercises 25–28 of the Supplementary Exercises for Chapters 5–8.) EXAMPLE 5 The Only Group of Order 255 is Z255 Let G be a group of order 255 5 3 ? 5 ? 17, and let H be a Sylow 17-subgroup of G. By Sylow’s Third Theorem, H is the only Sylow 17-subgroup of G, so N(H) 5 G. By Example 15 in Chapter 10, |N(H)/C(H)| divides |Aut(H)| 5 |Aut(Z17)|. By Theorem 6.5, |Aut(Z17)| 5 |U(17)| 5 16. Since |N(H)/C(H)| must divide 255 and 16, we have |N(H)/C(H)| 5 1. Thus, C(H) 5 G. This means that every element of G commutes with every element of H, and, therefore, H # Z(G). Thus, 17 divides |Z(G)|, which in turn divides 255. So |Z(G)| is equal to 17, 51, 85, or 255 and |G/Z(G)| is equal to 15, 5, 3, or 1. But the only groups of order 15, 5, 3, or 1 are the cyclic ones, so we know that G/Z(G) is cyclic. Now the “G/Z theorem” (Theorem 9.3) shows that G is Abelian, and the Fundamental Theorem of Finite Abelian Groups tells us that G is cyclic.

Exercises I have always grown from my problems and challenges, from the things that don’t work out. That’s when I’ve really learned. CAROL BURNETT

1. Show that conjugacy is an equivalence relation on a group. 2. Calculate all conjugacy classes for the quaternions (see Exercise 4, Supplementary Exercises for Chapters 1–4).

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3. Show that the function T defined in the proof of Theorem 24.1 is well defined, one-to-one, and maps the set of left cosets onto the conjugacy class of a. 4. Show that cl(a) 5 {a} if and only if a [ Z(G). 5. If |G| 5 36 and G is non-Abelian, prove that G has more than one Sylow 2-subgroup or more than one Sylow 3-subgroup. 6. Exhibit a Sylow 2-subgroup of S4. Describe an isomorphism from this group to D4. 7. Suppose that G is a group of order 48. Show that the intersection of any two distinct Sylow 2-subgroups of G has order 8. 8. Find all the Sylow 3-subgroups of S4. 9. Let K be a Sylow p-subgroup of a finite group G. Prove that if x [ N(K) and the order of x is a power of p, then x [ K. (This exercise is referred to in this chapter.) 10. If a group of order 52 ? 7 ? 11 has more than one Sylow 5-subgroup, exactly how many does it have? 11. Suppose that G is a group and ZGZ 5 p nm, where p is prime and p . m. Prove that a Sylow p-subgroup of G must be normal in G. 12. Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow p-subgroup of G contained in N(H). 13. Suppose that G is a group of order 168. If G has more than one Sylow 7-subgroup, exactly how many does it have? 14. Show that every group of order 56 has a proper nontrivial normal subgroup. 15. What is the smallest composite (that is, nonprime and greater than 1) integer n such that there is a unique group of order n? 16. Let G be a noncyclic group of order 21. How many Sylow 3subgroups does G have? 17. Prove that a noncyclic group of order 21 must have 14 elements of order 3. 18. How many Sylow 5-subgroups of S5 are there? Exhibit two. 19. How many Sylow 3-subgroups of S5 are there? Exhibit five. 20. Prove that a group of order 175 is Abelian. 21. Let H be a subgroup of a group G. Prove that the number of conjugates of H in G is |G:N(H)|. Hint: Mimic the proof of Theorem 24.1. (This exercise is referred to in this chapter and in Chapter 25.) 22. Generalize the argument given in Example 3 to obtain a theorem about groups of order p2q, where p and q are distinct primes. 23. What is the smallest possible odd integer that can be the order of a non-Abelian group?

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24. Prove that a group of order 375 has a subgroup of order 15. 25. Without using Theorem 24.6, prove that a group of order 15 is cyclic. (This exercise is referred to in the discussion about groups of order 30.) 26. Prove that a group of order 105 contains a subgroup of order 35. 27. Prove that a group of order 595 has a normal Sylow 17-subgroup. 28. Let G be a group of order 60. Show that G has exactly four elements of order 5 or exactly 24 elements of order 5. Which of these cases holds for A5? 29. Show that the center of a group of order 60 cannot have order 4. 30. Suppose that G is a group of order 60 and G has a normal subgroup N of order 2. Show that a. G has normal subgroups of orders 6, 10, and 30. b. G has subgroups of orders 12 and 20. c. G has a cyclic subgroup of order 30. 31. Let G be a group of order 60. If the Sylow 3-subgroup is normal, show that the Sylow 5-subgroup is normal. 32. Show that if G is a group of order 168 that has a normal subgroup of order 4, then G has a normal subgroup of order 28. 33. Suppose that p is prime and |G| 5 pn. Show that G has normal subgroups of order pk for all k between 1 and n (inclusive). 34. Suppose that G is a group of order pn, where p is prime, and G has exactly one subgroup for each divisor of pn. Show that G is cyclic. 35. Suppose that p is prime and |G| 5 pn. If H is a proper subgroup of G, prove that N(H) . H. (This exercise is referred to in Chapter 25.) 36. Suppose that G is a finite group and that all its Sylow subgroups are normal. Show that G is a direct product of its Sylow subgroups. 37. Let G be a finite group and let H be a normal Sylow p-subgroup of G. Show that a(H) 5 H for all automorphisms a of G. 38. If H is a normal subgroup of a finite group G and |H| 5 pk for some prime p, show that H is contained in every Sylow p-subgroup of G. 39. Let H and K denote a Sylow 3-subgroup and a Sylow 5-subgroup of a group, respectively. Suppose that |H| 5 3 and |K| 5 5. If 3 divides |N(K)|, show that 5 divides |N(H)|. 40. Let G be a group of order p2q2, where p and q are distinct primes, q B p2 2 1, and p B q2 2 1. Prove that G is Abelian. List three pairs of primes that satisfy these conditions.

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41. Let H be a normal subgroup of a group G. Show that H is the union of the conjugacy classes of the elements of H. Is this true when H is not normal in G? 42. Let p be prime. If the order of every element of a finite group G is a power of p, prove that |G| is a power of p. (Such a group is called a p-group.) 43. For each prime p, prove that all Sylow p-subgroups of a finite group are isomorphic. 44. Suppose that K is a normal subgroup of a finite group G and S is a Sylow p-subgroup of G. Prove that K > S is a Sylow psubgroup of K. 45. If G is a group of odd order and x [ G, show that x21 is not in cl(x). 46. Determine the groups of order 45. 47. Show that there are at most three nonisomorphic groups of order 21. 48. Prove that if H is a normal subgroup of index p2 where p is prime, then G9 # H (see Exercise 3 in the Supplementary Exercises for Chapters 5–8 for a description of G9). 49. Show that Z2 is the only group that has exactly two conjugacy classes. 50. If H is a finite subgroup of a group G and x [ G, prove that 0N(H) 0 5 0N(xHx 21 ) 0 . 51. Let G be a group with ZGZ 5 595 5 5 ? 7 ? 17. Show that the Sylow 5-subgroup of G is normal in G and is contained in Z(G). 52. What is the probability that a randomly selected element from D4 commutes with V? 53. Prove that if x and y are in the same conjugacy class of a group, then |C(x)| 5 |C(y)|. (This exercise is referred to in the discussion on the probability that two elements from a group commute.) 54. Let G be a finite group and let a [ G. Express the probability that a randomly selected element from G commutes with a in terms of orders of subgroups of G. 55. Find Pr(D4), Pr(S3), and Pr(A4). 56. Prove that Pr(G % H) 5 Pr(G) ? Pr(H). 57. Let R be a finite noncommutative ring. Show that the probability that two randomly chosen elements from R commute is at most 58 . [Hint: Mimic the group case and use the fact that the additive group R/C(R) is not cyclic.]

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Computer Exercise 1. Use the website http://people.csse.uwa.edu.au/gordon/remote/ cubcay/ to look up the number of groups of order 4p, where p is an odd prime up to 37. Make a conjecture about the exact number of such groups.

References 1. W. H. Gustafson, “What Is the Probability That Two Group Elements Commute?” The American Mathematical Monthly 80 (1973): 1031–1034. 2. Desmond MacHale, “How Commutative Can a Non-Commutative Group Be?” The Mathematical Gazette 58 (1974): 199–202. 3. E. A. O’Brien, “The Groups of Order Dividing 256,” Bulletin of the Australian Mathematical Society 39 (1989): 159–160. 4. H. Paley and P. Weichsel, A First Course in Abstract Algebra, New York: Holt, Rinehart & Winston, 1966.

Suggested Readings J. A. Gallian and D. Moulton, “When Is Zn the Only Group of Order n?” Elemente der Mathematik 48 (1993): 118–120. It is shown that Zn is the only group of order n if and only if n and f(n) are relatively prime. The article can be downloaded at http://www .d.umn.edu/~jgallian/pq.pdf. W. H. Gustafson, “What Is the Probability That Two Group Elements Commute?” The American Mathematical Monthly 80 (1973): 1031–1034. This paper is concerned with the problem posed in the title. It is shown that for all finite non-Abelian groups and certain infinite non-Abelian groups, the probability that two elements from a group commute is at most 5/8. The paper concludes with several exercises. Desmond MacHale, “Commutativity in Finite Rings,” The American Mathematical Monthly 83 (1976): 30–32. In this easy-to-read paper, it is shown that the probability that two elements from a finite noncommutative ring commute is at most 5/8. Several properties of Pr(G) when G is a finite group are stated. For example, if H # G, then Pr(G) # Pr(H). Also, there is no group G such that 7/16 , Pr(G) , 1/2.

Ludwig Sylow Sylow’s Theorem is 100 years old. In the course of a century this remarkable theorem has been the basis for the construction of numerous theories. L. A. SHEMETKOV

LUDWIG SYLOW (pronounced “SEE-loe”) was born on December 12, 1832, in Christiania (now Oslo), Norway. While a student at Christiania University, Sylow won a gold medal for competitive problem solving. In 1855, he became a high school teacher; despite the long hours required by his teaching duties, Sylow found time to study the papers of Abel. During the school year 1862–1863, Sylow received a temporary appointment at Christiania University and gave lectures on Galois theory and permutation groups. Among his students that year was the great mathematician Sophus Lie (pronounced “Lee”), after whom Lie algebras and Lie groups are named. From 1873 to 1881, Sylow, with some help from Lie, prepared a new edition of Abel’s works. In 1902, Sylow and Elling Holst published Abel’s correspondence. Sylow’s great discovery, Sylow’s Theorem, came in 1872. Upon learning of

Sylow’s result, C. Jordan called it “one of the essential points in the theory of permutations.” The result took on greater importance when the theory of abstract groups flowered in the late 19th century and early 20th century. In 1869, Sylow was offered a professorship at Christiania University but turned it down. Upon Sylow’s retirement from high school teaching at age 65, Lie mounted a successful campaign to establish a chair for Sylow at Christiania University. Sylow held this position until his death on September 7, 1918. To find more information about Sylow, visit: http://www-groups.dcs.st-and .ac.uk/~history

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25 Finite Simple Groups

It is a widely held opinion that the problem of classifying finite simple groups is close to a complete solution. This will certainly be one of the great achievements of mathematics of this century. NATHAN JACOBSON

Historical Background We now come to the El Dorado of finite group theory—the simple groups. Simple group theory is a vast and difficult subject; we call it the El Dorado of group theory because of the enormous effort put forth by hundreds of mathematicians over many years to discover and classify all finite simple groups. Let’s begin our discussion with the definition of a simple group and some historical background. Definition Simple Group

A group is simple if its only normal subgroups are the identity subgroup and the group itself.

The notion of a simple group was introduced by Galois about 180 years ago. The simplicity of A5, the group of even permutations on five symbols, played a crucial role in his proof that there is not a solution by radicals of the general fifth-degree polynomial (that is, there is no “quintic formula”). But what makes simple groups important in the theory of groups? They are important because they play a role in group theory somewhat analogous to that of primes in number theory or the elements in chemistry; that is, they serve as the building blocks for all groups. These building blocks may be determined in the following way. Given a finite group G, choose a proper normal subgroup G1 of G 5 G0 of largest order. Then the factor group G0/G1 is simple, and we next choose a proper normal subgroup G2 of G1 of largest order. Then G1/G2 is also simple, and we continue in this fashion until we arrive at Gn 5 {e}. The simple groups G0 /G1, G1/G2, . . . , Gn21/Gn are called the composition factors of G. More than 100 years ago, Jordan and Hölder 420

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proved that these factors are independent of the choices of the normal subgroups made in the process described. In a certain sense, a group can be reconstructed from its composition factors, and many of the properties of a group are determined by the nature of its composition factors. This and the fact that many questions about finite groups can be reduced (by induction) to questions about simple groups make clear the importance of determining all finite simple groups. Just which groups are the simple ones? The Abelian simple groups are precisely Zn, where n 5 1 or n is prime. This follows directly from the corollary in Chapter 11. In contrast, it is extremely difficult to describe the non-Abelian simple groups. The best we can do here is to give a few examples and mention a few words about their discovery. It was Galois in 1831 who first observed that An is simple for all n $ 5. The next discoveries were made by Jordan in 1870, when he found four infinite families of simple matrix groups over the field Zp, where p is prime. One such family is the factor group SL(n, Zp)/Z(SL(n, Zp)), except when n 5 2 and p 5 2 or p 5 3. Between the years 1892 and 1905, the American mathematician Leonard Dickson (see Chapter 22 for a biography) generalized Jordan’s results to arbitrary finite fields and discovered several new infinite families of simple groups. About the same time, it was shown by G. A. Miller and F. N. Cole that a family of five groups first described by E. Mathieu in 1861 were in fact simple groups. Since these five groups were constructed by ad hoc methods that did not yield infinitely many possibilities, like An or the matrix groups over finite fields, they were called “sporadic.” The next important discoveries came in the 1950s. In that decade, many new infinite families of simple groups were found, and the initial steps down the long and winding road that led to the complete classification of all finite simple groups were taken. The first step was Richard Brauer’s observation that the centralizer of an element of order 2 was an important tool for studying simple groups. A few years later, John Thompson, in his Ph.D. dissertation, introduced the crucial idea of studying the normalizers of various subgroups of prime-power order. In the early 1960s came the momentous Feit-Thompson Theorem, which says that a non-Abelian simple group must have even order. This property was first conjectured around 1900 by one of the pioneers of modern group-theoretic methods, the Englishman William Burnside (see Chapter 29 for a biography). The proof of the Feit-Thompson Theorem filled an entire issue of a journal [1], 255 pages in all (see Figure 25.1). This result provided the impetus to classify the finite simple groups—that is, a program to discover all finite simple groups and prove that there are no more to be found. Throughout the 1960s, the

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Oh, what are the orders of all simple groups? I speak of the honest ones, not of the loops. It seems that old Burnside their orders has guessed Except for the cyclic ones, even the rest. CHORUS: Finding all groups that are simple is no simple task. Groups made up with permutes will produce some more: For An is simple, if n exceeds 4. Then, there was Sir Matthew who came into view Exhibiting groups of an order quite new. Still others have come on to study this thing. Of Artin and Chevalley now we shall sing. With matrices finite they made quite a list The question is: Could there be others they’ve missed? Suzuki and Ree then maintained it’s the case

That these methods had not reached the end of the chase. They wrote down some matrices, just four by four. That made up a simple group. Why not make more? And then came the opus of Thompson and Feit Which shed on the problem remarkable light. A group, when the order won’t factor by two, Is cyclic or solvable. That’s what is true. Suzuki and Ree had caused eyebrows to raise, But the theoreticians they just couldn’t faze. Their groups were not new: if you added a twist, You could get them from old ones with a flick of the wrist. Still, some hardy souls felt a thorn in their side.

Figure 25.1

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For the five groups of Mathieu all reason defied; Not An, not twisted, and not Chevalley, They called them sporadic and filed them away. Are Mathieu groups creatures of heaven or hell? Zvonimir Janko determined to tell. He found out that nobody wanted to know: The masters had missed 1 7 5 5 6 0.

423

(And twelve or more sprouted, to greet the new age.) By Janko and Conway and Fischer and Held McLaughlin, Suzuki, and Higman, and Sims. No doubt you noted the last lines don’t rhyme. Well, that is, quite simply, a sign of the time. There’s chaos, not order, among simple groups; And maybe we’d better go back to the loops.

The floodgates were opened! New groups were the rage!

methods introduced in the Feit-Thompson proof were generalized and improved with great success by many mathematicians. Moreover, between 1966 and 1975, 19 new sporadic simple groups were discovered. Despite many spectacular achievements, research in simple group theory in the 1960s was haphazard, and the decade ended with many people believing that the classification would never be completed. (The pessimists feared that the sporadic simple groups would foil all attempts. The anonymously written “song” in Figure 25.1 captures the spirit of the times.) Others, more optimistic, were predicting that it would be accomplished in the 1990s. The 1970s began with Thompson receiving the Fields Medal for his fundamental contributions to simple group theory. This honor is among the highest forms of recognition that a mathematician can receive (more information about the Fields Medal is given near the end of this chapter). Within a few years, three major events took place that ultimately led to the classification. First, Thompson published what is regarded as the single most important paper in simple group theory—the N-group paper. Here, Thompson introduced many fundamental techniques and supplied a model for the classification of a broad family of simple groups. Second, Daniel Gorenstein produced an elaborate outline for the classification, which he delivered in a series of lectures at the University of Chicago in 1972. Here a program for the overall proof was laid out. The army of researchers now had a battle plan and a commander-in-chief. But this army still needed more and better

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weapons. Thus came the third critical development: the involvement of Michael Aschbacher. In a dazzling series of papers, Aschbacher combined his own insight with the methods of Thompson, which had been generalized throughout the 1960s, and a geometric approach pioneered by Bernd Fischer to achieve one brilliant result after another in rapid succession. In fact, so much progress was made by Aschbacher and others that by 1976, it was clear to nearly everyone involved that enough techniques had been developed to complete the classification. Only details remained. The 1980s were ushered in with Aschbacher following in the footsteps of Feit and Thompson by winning the American Mathematical Society’s Cole Prize in algebra (see the last section of this chapter). A week later, Robert L. Griess made the spectacular announcement that he had constructed the “Monster.”† The Monster is the largest of the sporadic simple groups. In fact, it has vastly more elements than there are atoms on the earth! Its order is 808,017,424,794,512,875,886,459,904,961,710,757,005,754, 368,000,000,000 (hence, the name). This is approximately 8 3 1053. The Monster is a group of rotations in 196,883 dimensions. Thus, each element can be expressed as a 196,883 3 196,883 matrix. At the annual meeting of the American Mathematical Society in 1981, Gorenstein announced that the “Twenty-five Years’ War” to classify all the finite simple groups was over. Group theorists at long last had a list of all finite simple groups and a proof that the list was complete. The proof was spread out over hundreds of papers—both published and unpublished— and ran more than 10,000 pages in length. Because of the proof’s extreme length and complexity, and the fact that some key parts of it had not been published, there was some concern in the mathematics community that the classification was not a certainty. By the end of the decade, group theorists had concluded that there was indeed a gap in the unpublished work that would be difficult to rectify. In the mid-1990s, Aschbacher and Stephen Smith began work on this problem. In 2004, at the annual meeting of the American Mathematical Society, Aschbacher announced that he and Smith had completed the classification. Their paper is about 1200 pages in length. Aschbacher concluded his remarks by saying that he would not bet his house that the proof is now error-free. Several people who played a central role in the classification are working on a “second generation” proof that will be much shorter and more comprehensible.

†The

name was coined by John H. Conway.

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Nonsimplicity Tests In view of the fact that simple groups are the building blocks for all groups, it is surprising how scarce the non-Abelian simple groups are. For example, A5 is the only one whose order is less than 168; there are only five non-Abelian simple groups of order less than 1000 and only 56 of order less than 1,000,000. In this section, we give a few theorems that are useful in proving that a particular integer is not the order of a non-Abelian simple group. Our first such result is an easy arithmetic test that comes from combining Sylow’s Third Theorem and the fact that groups of prime-power order have nontrivial centers. Theorem 25.1 Sylow Test for Nonsimplicity Let n be a positive integer that is not prime, and let p be a prime divisor of n. If 1 is the only divisor of n that is equal to 1 modulo p, then there does not exist a simple group of order n.

PROOF If n is a prime-power, then a group of order n has a nontrivial center and, therefore, is not simple. If n is not a prime-power, then every Sylow subgroup is proper, and, by Sylow’s Third Theorem, we know that the number of Sylow p-subgroups of a group of order n is equal to 1 modulo p and divides n. Since 1 is the only such number, the Sylow p-subgroup is unique, and therefore, by the corollary to Sylow’s Third Theorem, it is normal. How good is this test? Well, applying this criterion to all the nonprime integers between 1 and 200 would leave only the following integers as possible orders of finite non-Abelian simple groups: 12, 24, 30, 36, 48, 56, 60, 72, 80, 90, 96, 105, 108, 112, 120, 132, 144, 150, 160, 168, 180, and 192. (In fact, computer experiments have revealed that for large intervals, say, 500 or more, this test eliminates more than 90% of the nonprime integers as possible orders of simple groups. See [2] for more on this.) Our next test rules out 30, 90, and 150. Theorem 25.2 2 ? Odd Test An integer of the form 2 ? n, where n is an odd number greater than 1, is not the order of a simple group.

PROOF Let G be a group of order 2n, where n is odd and greater than 1. Recall from the proof of Cayley’s Theorem (Theorem 6.1)

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that the mapping g S Tg is an isomorphism from G to a permutation group on the elements of G [where Tg(x) 5 gx for all x in G]. Since |G| 5 2n, Cauchy’s Theorem guarantees that there is an element g in G of order 2. Then, when the permutation Tg is written in disjoint cycle form, each cycle must have length 1 or 2; otherwise, |g| 2 2. But Tg can contain no 1-cycles, because the 1-cycle (x) would mean x 5 Tg(x) 5 gx, so g 5 e. Thus, in cycle form, Tg consists of exactly n transpositions, where n is odd. Therefore, Tg is an odd permutation. This means that the set of even permutations in the image of G is a normal subgroup of index 2. (See Exercise 19 in Chapter 5 and Exercise 7 in Chapter 9.) Hence, G is not simple. The next theorem is a broad generalization of Cayley’s Theorem. We will make heavy use of its two corollaries. Theorem 25.3 Generalized Cayley Theorem Let G be a group and let H be a subgroup of G. Let S be the group of all permutations of the left cosets of H in G. Then there is a homomorphism from G into S whose kernel lies in H and contains every normal subgroup of G that is contained in H.

PROOF For each g [ G, define a permutation Tg of the left cosets of H by Tg(xH) 5 gxH. As in the proof of Cayley’s Theorem, it is easy to verify that the mapping of a:g S Tg is a homomorphism from G into S. Now, if g [ Ker a, then Tg is the identity map, so H 5 Tg(H) 5 gH, and, therefore, g belongs to H. Thus, Ker a # H. On the other hand, if K is normal in G and K # H, then for any k [ K and any x in G, there is an element k9 in K such that kx 5 xk9. Thus, Tk(xH) 5 kxH 5 xk9H 5 xH and, therefore, Tk is the identity permutation. This means that k [ Ker a. We have proved, then, that every normal subgroup of G contained in H is also contained in Ker a. As a consequence of Theorem 25.3, we obtain the following very powerful arithmetic test for nonsimplicity. Corollary 1 Index Theorem If G is a finite group and H is a proper subgroup of G such that |G| does not divide |G:H|!, then H contains a nontrivial normal subgroup of G. In particular, G is not simple.

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PROOF Let a be the homomorphism given in Theorem 25.3. Then Ker a is a normal subgroup of G contained in H, and G/Ker a is isomorphic to a subgroup of S. Thus, |G/Ker a| 5 |G|/|Ker a| divides |S| 5 |G:H|!. Since |G| does not divide |G:H|!, the order of Ker a must be greater than 1. Corollary 2 Embedding Theorem If a finite non-Abelian simple group G has a subgroup of index n, then G is isomorphic to a subgroup of An.

PROOF Let H be the subgroup of index n, and let Sn be the group of all permutations of the n left cosets of H in G. By the Generalized Cayley Theorem, there is a nontrivial homomorphism from G into Sn. Since G is simple and the kernel of a homomorphism is a normal subgroup of G, we see that the mapping from G into Sn is one-to-one, so that G is isomorphic to some subgroup of Sn. Recall from Exercise 19 in Chapter 5 that any subgroup of Sn consists of even permutations only or half even and half odd. If G were isomorphic to a subgroup of the latter type, the even permutations would be a normal subgroup of index 2 (see Exercise 7 in Chapter 9), which would contradict the fact that G is simple. Thus, G is isomorphic to a subgroup of An. Using the Index Theorem with the largest Sylow subgroup for H reduces our list of possible orders of non-Abelian simple groups still further. For example, let G be any group of order 80 5 16 ? 5. We may choose H to be a subgroup of order 16. Since 80 is not a divisor of 5!, there is no simple group of order 80. The same argument applies to 12, 24, 36, 48, 96, 108, 160, and 192, leaving only 56, 60, 72, 105, 112, 120, 132, 144, 168, and 180 as possible orders of non-Abelian simple groups up to 200. Let’s consider these orders. Quite often we may use a counting argument to eliminate an integer. Consider 56. By Sylow’s Theorem, we know that a simple group of order 56 5 8 ? 7 would contain eight Sylow 7-subgroups and seven Sylow 2-subgroups. Now, any two Sylow p-subgroups that have order p must intersect in only the identity. So the union of the eight Sylow 7-subgroups yields 48 elements of order 7, and the union of any two Sylow 2-subgroups gives at least 8 1 8 2 4 5 12 new elements. But there are only 56 elements in all. This contradiction shows that there is not a simple group of order 56. An analogous argument also eliminates the integers 105 and 132. So, our list of possible orders of non-Abelian simple groups up to 200 is down to 60, 72, 112, 120, 144, 168, and 180. Of these, 60 and

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168 do correspond to simple groups. The others can be eliminated with a bit of razzle-dazzle. The easiest case to handle is 112 5 24 ? 7. Suppose there were a simple group G of order 112. A Sylow 2-subgroup of G must have index 7. So, by the Embedding Theorem, G is isomorphic to a subgroup of A7. But 112 does not divide |A7|, which is a contradiction. Another easy case is 72 5 23 ? 32. Recall from Exercise 21 in Chapter 24 that the number of Sylow p-subgroups of G is np 5 |G: N(H)|, where H is any Sylow p-subgroup of G. Now for a simple group G of order 72, Sylow’s Third Theorem gives that n3|8 and n3 mod 3 5 1. So n3 5 4. Since |G| 5 72 does not divide |G: N(H)|! 5 4! 5 24, the Index Theorem gives that G is not simple. Next consider the possibility of a simple group G of order 144 5 9 ? 16. By the Sylow theorems, we know that n3 5 4 or n3 5 16 and n2 $ 3. The Index Theorem rules out the case where n3 5 4, so we know that there are 16 Sylow 3-subgroups. Now, if every pair of Sylow 3-subgroups had only the identity in common, the 16 ? 8 1 1 5 129 elements from these subgroups plus at least 16 more elements from two of the Sylow 2-subgroups results in more than 144 elements. So, let H and H9 be a pair of Sylow 3-subgroups whose intersection has order 3. Then H > H9 is a subgroup of both H and H9 and, by the corollary to Theorem 24.2 (or by Exercise 35 in Chapter 24), we see that N(H > H9) must contain both H and H9 and, therefore, the set HH9. (HH9 need not be a subgroup.) Thus, |N(H > H9)| $ |HH9| 5

0 H 0 0 Hr 0 9.9 5 27. 5 0 H d Hr 0 3

Now, we have three arithmetic conditions on k 5 |N(H > H9)|. We know that 9 divides k; k divides 144; and k $ 27. Clearly, then, k $ 36, and so |G : N(H > H9)| # 4. The Index Theorem now gives us the desired contradiction. Finally, suppose that G is a non-Abelian simple group of order 180 5 22 ? 32 ? 5. Then n5 5 6 or n5 5 36 and n3 5 10. First, assume that n5 5 36. Then G has 36 ? 4 5 144 elements of order 5. Now, if each pair of the Sylow 3-subgroups intersects in only the identity, then there are 80 more elements in the group, which is a contradiction. So, we may assume that there are two Sylow 3-subgroups L3 and L 39 whose intersection has order 3. Then, as was the case for order 144, we have 9.9 |N(L3 > L 39)| $ |L3L 39| 5 5 27. 3 Thus, |N(L3 > L 39)| 5 9 ? k,

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where k $ 3 and k divides 20. Clearly, then, |N(L3 > L 39)| $ 36 and therefore |G:N(L3 > L 39)| # 5. The Index Theorem now gives us another contradiction. Hence, we may assume that n5 5 6. In this case, we let H be the normalizer of a Sylow 5-subgroup of G. By Sylow’s Third Theorem, we have 6 5 |G:H|, so that |H| 5 30. In Chapter 24, we proved that every group of order 30 has an element of order 15. On the other hand, since n5 5 6, G has a subgroup of index 6 and the Embedding Theorem tells us that G is isomorphic to a subgroup of A6. But A6 has no element of order 15. (See Exercise 7 in Chapter 5.) Unfortunately, the argument for 120 is fairly long and complicated. However, no new techniques are required to do it. We leave this as an exercise (Exercise 17). Some hints are given in the answer section.

The Simplicity of A5 Once 120 has been disposed of, we will have shown that the only integers between 1 and 200 that can be the orders of non-Abelian simple groups are 60 and 168. For completeness, we will now prove that A5, which has order 60, is a simple group. A similar argument can be used to show that the factor group SL(2, Z7)/Z(SL(2, Z7)) is a simple group of order 168. [This group is denoted by PSL(2, Z7).] If A5 had a nontrivial proper normal subgroup H, then |H| is equal to 2, 3, 4, 5, 6, 10, 12, 15, 20, or 30. By Exercise 43 in Chapter 5, A5 has 24 elements of order 5, 20 elements of order 3, and no elements of order 15. Now, if |H| is equal to 3, 6, 12, or 15, then |A5/H| is relatively prime to 3, and by Exercise 59 in Chapter 9, H would have to contain all 20 elements of order 3. If |H| is equal to 5, 10, or 20, then |A5/H| is relatively prime to 5, and, therefore, H would have to contain the 24 elements of order 5. If |H| 5 30, then |An5 /H| is relatively prime to both 3 and 5, and so H would have to contain all the elements of orders 3 and 5. Finally, if |H| 5 2 or |H| 5 4, then |A5/H| 5 30 or |A5/H| 5 15. But we know from our results in Chapter 24 that any group of order 30 or 15 has an element of order 15. However, since A5 contains no such element, neither does A5/H. This proves that A5 is simple. The simplicity of A5 was known to Galois in 1830, although the first formal proof was done by Jordan in 1870. A few years later, Felix Klein showed that the group of rotations of a regular icosahedron is

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simple and, therefore, isomorphic to A5 (see Exercise 27). Since then it has frequently been called the icosahedral group. Klein was the first to prove that there is a simple group of order 168. The problem of determining which integers in a certain interval are possible orders for finite simple groups goes back to 1892, when Hölder went up to 200. His arguments for the integers 144 and 180 alone used up 10 pages. By 1975, this investigation had been pushed to well beyond 1,000,000. See [3] for a detailed account of this endeavor. Of course, now that all finite simple groups have been classified, this problem is merely a historical curiosity.

The Fields Medal Among the highest awards for mathematical achievement is the Fields Medal. Two to four such awards are bestowed at the opening session of the International Congress of Mathematicians, held once every four years. Although the Fields Medal is considered by many mathematicians to be the equivalent of the Nobel Prize, there are great differences between these awards. Besides the huge disparity in publicity and monetary value associated with the two honors, the Fields Medal is restricted to those under 40 years of age.† This tradition stems from John Charles Fields’s stipulation, in his will establishing the medal, that the awards should be “an encouragement for further achievement.” This restriction precluded Andrew Wiles from winning the Fields Medal for his proof of Fermat’s Last Theorem. More details about the Fields Medal can be found at www .wikipedia.com.

The Cole Prize Approximately every five years, beginning in 1928, the American Mathematical Society awards one or two Cole Prizes for research in algebra and one or two Cole Prizes for research in algebraic number theory. The prize was founded in honor of Frank Nelson Cole on the occasion of his retirement as secretary of the American Mathematical Society. In view of the fact that Cole was one of the first people interested in simple groups, it is interesting to note that no fewer than six †“Take

the sum of human achievement in action, in science, in art, in literature— subtract the work of the men above forty, and while we should miss great treasures, even priceless treasures, we would practically be where we are today. . . . The effective, moving, vitalizing work of the world is done between the ages of twenty-five and forty.” Sir William Osler (1849–1919), Life of Sir William Osler, vol. I, chap. 24 (The Fixed Period).

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recipients of the prize—Dickson, Chevalley, Brauer, Feit, Thompson, and Aschbacher—have made fundamental contributions to simple group theory at some time in their careers.

Exercises If you don’t learn from your mistakes, there’s no sense making them. HERBERT V. PROCHNOW

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

14. 15. 16. 17. 18. 19.

20.

Prove that there is no simple group of order 210 5 2 ? 3 ? 5 ? 7. Prove that there is no simple group of order 280 5 23 ? 5 ? 7. Prove that there is no simple group of order 216 5 23 ? 33. Prove that there is no simple group of order 300 5 22 ? 3 ? 52. Prove that there is no simple group of order 525 5 3 ? 52 ? 7. Prove that there is no simple group of order 540 5 22 ? 33 ? 5. Prove that there is no simple group of order 528 5 24 ? 3 ? 11. Prove that there is no simple group of order 315 5 32 ? 5 ? 7. Prove that there is no simple group of order 396 5 22 ? 32 ? 11. Prove that there is no simple group of order n, where 201 # n # 235. Without using the Generalized Cayley Theorem or its corollaries, prove that there is no simple group of order 112. Without using the “2 ? odd” test, prove that there is no simple group of order 210. You may have noticed that all the “hard integers” are even. Choose three odd integers between 200 and 1000. Show that none of these is the order of a simple group unless it is prime. Show that there is no simple group of order pqr, where p, q, and r are primes ( p, q, and r need not be distinct). Show that A5 does not contain a subgroup of order 30, 20, or 15. Show that S5 does not contain a subgroup of order 40 or 30. Prove that there is no simple group of order 120 5 23 ? 3 ? 5. (This exercise is referred to in this chapter.) Prove that if G is a finite group and H is a proper normal subgroup of largest order, then G/H is simple. Suppose that H is a subgroup of a finite group G and that |H| and (|G:H| 2 1)! are relatively prime. Prove that H is normal in G. What does this tell you about a subgroup of index 2 in a finite group? Suppose that p is the smallest prime that divides |G|. Show that any subgroup of index p in G is normal in G.

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21. Prove that the only nontrivial proper normal subgroup of S5 is A5. (This exercise is referred to in Chapter 32.) 22. Prove that a simple group of order 60 has a subgroup of order 6 and a subgroup of order 10. 23. Show that PSL(2, Z7) 5 SL(2, Z7)/Z(SL(2, Z7)), which has order 168, is a simple group. (This exercise is referred to in this chapter.) 24. Show that the permutations (12) and (12345) generate S5. 25. Suppose that a subgroup H of S5 contains a 5-cycle and a 2-cycle. Show that H 5 S5. (This exercise is referred to in Chapter 32.) 26. Suppose that G is a finite simple group and contains subgroups H and K such that |G:H| and |G:K| are prime. Show that |H| 5 |K|. 27. Show that (up to isomorphism) A5 is the only simple group of order 60. (This exercise is referred to in this chapter.) 28. Prove that a simple group cannot have a subgroup of index 4. 29. Prove that there is no simple group of order p2q, where p and q are odd primes and q . p. 30. If a simple group G has a subgroup K that is a normal subgroup of two distinct maximal subgroups, prove that K 5 5e6 . 31. Show that a finite group of even order that has a cyclic Sylow 2subgroup is not simple.

Computer Exercises They have computers, and they may have other weapons of mass destruction. JANET RENO

Software for the computer exercises in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This software uses a counter M to keep track of how many integers Theorem 25.1 eliminates in any given interval. Run the program for the following intervals: 1–100; 501–600; 5001–5100; 10,001–10,100. How does M seem to behave as the sizes of the integers grow? 2. This software uses a counter M to keep track of how many integers the Index Theorem eliminates in any given interval of integers. Run the program for the same intervals as in Exercise 1. How does M seem to behave as the sizes of the integers grow?

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References 1. W. Feit and J. G. Thompson, “Solvability of Groups of Odd Order,” Pacific Journal of Mathematics 13 (1963): 775–1029. 2. J. A. Gallian, “Computers in Group Theory,” Mathematics Magazine 49 (1976): 69–73. 3. J. A. Gallian, “The Search for Finite Simple Groups,” Mathematics Magazine 49 (1976): 163–179.

Suggested Readings G. Cornell, N. Pele, and M. Wage, “Simple Groups of Orders Less Than 1000,” Journal of Undergraduate Research 5 (1973): 77–86. In this charming article, three undergraduate students use slightly more theory than was given in this chapter to show that the only integers less than 1000 that could be orders of simple groups are 60, 168, 320, 504, 660, and 720. All but the last one are orders of simple groups. The proof that there is no simple group of order 720 is omitted because it is significantly beyond most undergraduates. K. David, “Using Commutators to Prove A5 Is Simple,” The American Mathematical Monthly 94 (1987): 775–776. This note gives an elementary proof that A5 is simple using commutators. J. A. Gallian, “The Search for Finite Simple Groups,” Mathematics Magazine 49 (1976): 163–179 (http://www.d.umn.edu/~jgallian/simple.pdf). A historical account is given of the search for finite simple groups. Martin Gardner, “The Capture of the Monster: A Mathematical Group with a Ridiculous Number of Elements,” Scientific American 242 (6) (1980): 20–32. This article gives an elementary introduction to groups and a discussion of simple groups, including the “Monster.” Daniel Gorenstein, “The Enormous Theorem,” Scientific American 253 (6) (1985): 104–115. You won’t find an article on a complex subject better written for the layperson than this one. Gorenstein, the driving force behind the classification, uses concrete examples, analogies, and nontechnical terms to make the difficult subject matter of simple groups accessible. Sandra M. Lepsi, “PSL(2, Z7) Is Simple, by Counting,” Pi Mu Epsilon Journal, Fall (1993): 576–578. The author shows that the group SL(2, Z7)/Z(SL(2, Z7)) of order 168 is simple using a counting argument.

Michael Aschbacher Fresh out of graduate school, he [Aschbacher] had just entered the field, and from that moment he became the driving force behind my program. In rapid succession he proved one astonishing theorem after another. Although there were many other major contributors to this final assault, Aschbacher alone was responsible for shrinking my projected 30-year timetable to a mere 10 years. DANIEL GORENSTEIN, Scientific American

MICHAEL ASCHBACHER was born on April 8, 1944, in Little Rock, Arkansas. Shortly after his birth, his family moved to Illinois, where his father was a professor of accounting and his mother was a high school English teacher. When he was nine years old, his family moved to East Lansing, Michigan; six years later, they moved to Los Angeles. After high school, Aschbacher enrolled at the California Institute of Technology. In addition to his schoolwork, he passed the first four actuary exams and was employed for a few years as an actuary, full-time in the summers and part-time during the academic year. Two of the Caltech mathematicians who influenced him were Marshall Hall and Donald Knuth. In his senior year, Aschbacher took abstract algebra but showed little interest in the course. Accordingly, he received a grade of C. In 1966, Aschbacher went to the University of Wisconsin for a Ph.D. degree. He

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completed his dissertation in 1969, and, after spending one year as an assistant professor at the University of Illinois, he returned to Caltech and quickly moved up to the rank of professor. Aschbacher’s dissertation work in the area of combinatorial geometries had led him to consider certain group-theoretic questions. Gradually, he turned his attention more and more to purely group-theoretic problems, particularly those bearing on the classification of finite simple groups. The 1980 Cole Prize Selection Committee said of one of his papers, “[It] lifted the subject to a new plateau and brought the classification within reach.” Aschbacher has been elected to the National Academy of Sciences, the American Academy of Sciences, and the vice presidency of the American Mathematical Society.

Daniel Gorenstein Gorenstein was one of the most influential mathematicians of the last few decades. MICHAEL ASCHBACHER, Notices of the American Mathematical Society 39 (1992): 1190

DANIEL GORENSTEIN was born in Boston on January 1, 1923. Upon graduating from Harvard in 1943 during World War II, Gorenstein was offered an instructorship at Harvard to teach mathematics to army personnel. After the war ended, he began graduate work at Harvard. He received his Ph.D. degree in 1951, working in algebraic geometry under Oscar Zariski. It was in his dissertation that he introduced the class of rings that is now named after him. In 1951, Gorenstein took a position at Clark University in Worcester, Massachusetts, where he stayed until moving to Northeastern University in 1964. From 1969 until his death on August 26, 1992, he was at Rutgers University. In 1957, Gorenstein switched from algebraic geometry to finite groups, learning the basic material from I. N. Herstein while collaborating with him over the next few years. A milestone in Gorenstein’s development as a group theorist came during 1960–1961, when he was invited to participate in a “Group Theory Year” at the University of Chicago.

It was there that Gorenstein, assimilating the revolutionary techniques then being developed by John Thompson, began his fundamental work that contributed to the classification of finite simple groups. Through his pioneering research papers, his dynamic lectures, his numerous personal contacts, and his influential book on finite groups, Gorenstein became the leader in the 25-year effort, by hundreds of mathematicians, that led to the classification of the finite simple groups. Among the honors received by Gorenstein are the Steele Prize from the American Mathematical Society and election to membership in the National Academy of Sciences and the American Academy of Arts and Sciences. To find more information about Gorenstein, visit: http://www-groups.dcs.st-and .ac.uk/~history/

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John Thompson There seemed to be no limit to his power. DANIEL GORENSTEIN

JOHN G. THOMPSON was born on October 13, 1932, in Ottawa, Kansas. In 1951, he entered Yale University as a divinity student but switched to mathematics in his sophomore year. In 1955, he began graduate school at the University of Chicago and obtained his Ph.D. degree four years later. After one year on the faculty at Harvard, Thompson returned to Chicago. He remained there until 1968, when he moved to Cambridge University in England. In 1993, Thompson accepted an appointment at the University of Florida. Thompson’s brilliance was evident early. In his dissertation, he verified a 50-year-old conjecture about finite groups possessing a certain kind of automorphism. (An article about his achievement appeared in The New York Times.) The novel methods Thompson used in his dissertation foreshadowed the revolutionary ideas he would later introduce in the Feit-Thompson paper and the classification of minimal simple groups (simple groups that contain no proper non-Abelian simple subgroups). The assimilation and extension of Thompson’s methods by others throughout the 1960s and 1970s ultimately

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led to the classification of finite simple groups. In the late 1970s, Thompson made significant contributions to coding theory, the theory of finite projective planes, and the theory of modular functions. His recent work on Galois groups is considered the most important in the field in the last half of the 20th century. Among Thompson’s many honors are the Cole Prize in algebra and the Fields Medal. He was elected to the National Academy of Sciences in 1967, the Royal Society of London in 1979, and the American Academy of Arts and Sciences in 1998. In 2000, President Clinton presented Thompson the National Medal of Science. In 2008 he was a co-winner of the $1,000,000 Abel Prize given by the Norwegian Academy of Science and Letters. To find more information about Thompson, visit: http://www-groups.dcs.st-and .ac.uk/~history/

Generators 26 and Relations One cannot escape the feeling that these mathematical formulae have an independent existence and an intelligence of their own, that they are wiser than we are, wiser even than their discoverers, that we get more out of them than we originally put into them. HENRICH HERTZ

Motivation In this chapter, we present a convenient way to define a group with certain prescribed properties. Simply put, we begin with a set of elements that we want to generate the group, and a set of equations (called relations) that specify the conditions that these generators are to satisfy. Among all such possible groups, we will select one that is as large as possible. This will uniquely determine the group up to isomorphism. To provide motivation for the theory involved, we begin with a concrete example. Consider D4, the group of symmetries of a square. Recall that R 5 R90 and H, a reflection across a horizontal axis, generate the group. Observe that R and H are related in the following ways: R4 5 H 2 5 (RH)2 5 R0

(the identity).

(1)

Other relations between R and H, such as HR 5 R3H and RHR 5 H, also exist, but they can be derived from those given in Equation (1). For example, (RH)2 5 R0 yields HR 5 R21H21, and R4 5 H 2 5 R0 yields R21 5 R3 and H21 5 H. So, HR 5 R3H. In fact, every relation between R and H can be derived from those given in Equation (1). Thus, D4 is a group that is generated by a pair of elements a and b subject to the relations a4 5 b2 5 (ab)2 5 e and such that all other relations between a and b can be derived from these relations. This last stipulation is necessary because the subgroup {R0, R180, H, V} of D4 is generated by the two elements a 5 R180 and b 5 H that satisfy the relations a4 5 b2 5 (ab)2 5 e. However, the “extra” relation a2 5 e satisfied by this subgroup cannot be derived from the original ones (since R9022 R0). It is natural to ask whether this description of D4 applies to some other group 437

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as well. The answer is no. Any other group generated by two elements a and b satisfying only the relations a4 5 b2 5 (ab)2 5 e, and those that can be derived from these relations, is isomorphic to D4. Similarly, one can show that the group Z4 % Z2 is generated by two elements a and b such that a4 5 b2 5 e and ab 5 ba, and any other relation between a and b can be derived from these relations. The purpose of this chapter is to show that this procedure can be reversed; that is, we can begin with any set of generators and relations among the generators and construct a group that is uniquely described by these generators and relations, subject to the stipulation that all other relations among the generators can be derived from the original ones.

Definitions and Notation We begin with some definitions and notation. For any set S 5 {a, b, c, . . .} of distinct symbols, we create a new set S21 5 {a21, b21, c21, . . .} by replacing each x in S by x21. Define the set W(S) to be the collection of all formal finite strings of the form x1x2 ? ? ? xk, where each xi [ S < S21. The elements of W(S) are called words from S. We also permit the string with no elements to be in W(S). This word is called the empty word and is denoted by e. We may define a binary operation on the set W(S) by juxtaposition; that is, if x1x2 ? ? ? xk and y1y2 ? ? ? yt belong to W(S), then so does x1x2 ? ? ? xk y1y2 ? ? ? yt. Observe that this operation is associative and the empty word is the identity. Also, notice that a word such as aa21 is not the identity, because we are treating the elements of W(S) as formal symbols with no implied meaning. At this stage we have everything we need to make a group out of W(S) except inverses. Here a difficulty arises, since it seems reasonable that the inverse of the word ab, say, should be b21a21. But abb21a21 is not the empty word! You may recall that we faced a similar obstacle long ago when we carried out the construction of the field of quotients of an integral domain. There we had formal symbols of the form a/b and we wanted the inverse of a/b to be b/a. But their product, ab/(ba), was a formal symbol that was not the same as the formal symbol 1/1, the identity. So, we proceed here as we did there—by way of equivalence classes. Definition Equivalence Classes of Words

For any pair of elements u and v of W(S), we say that u is related to v if v can be obtained from u by a finite sequence of insertions or deletions of words of the form xx21 or x21x, where x [ S.

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We leave it as an exercise to show that this relation is an equivalence relation on W(S). (See Exercise 1.) EXAMPLE 1 Let S 5 {a, b, c}. Then acc21b is equivalent to ab; aab21bbaccc21 is equivalent to aabac; the word a21aabb21a21 is equivalent to the empty word; and the word ca21b is equivalent to cc21caa21a21bbca21ac21b21. Note, however, that cac21b is not equivalent to ab.

Free Group Theorem 26.1 Equivalence Classes Form a Group Let S be a set of distinct symbols. For any word u in W(S), let u denote the set of all words in W(S) equivalent to u (that is, u is the equivalence class containing u). Then the set of all equivalence classes of elements of W(S) is a group under the operation u ? v 5 uv.

PROOF This proof is left to the reader. The group defined in Theorem 26.1 is called a free group on S. Theorem 26.2 shows why free groups are important. Theorem 26.2 The Universal Mapping Property Every group is a homomorphic image of a free group.

PROOF Let G be a group and let S be a set of generators for G. (Such a set exists, because we may take S to be G itself.) Now let F be the free group on S. Unfortunately, since our notation for any word in W(S) also denotes an element of G, we have created a notational problem for ourselves. So, to distinguish between these two cases, we will denote the word x1x2 ? ? ? xn in W(S) by (x1x2 ? ? ? xn)F and the product x1x2 ? ? ? xn in G by (x1x2 ? ? ? xn)G. As before, x1x2 . . . xn denotes the equivalence class in F containing the word (x1x2 ? ? ? xn)F in W(S). Notice that x1x2 . . . xn and (x1x2 ? ? ? xn)G are entirely different elements, since the operations on F and G are different. Now consider the mapping from F into G given by f(x1x2 . . . xn) 5 (x1x2 ? ? ? xn)G.

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[All we are doing is taking a product in F and viewing it as a product in G. For example, if G is the cyclic group of order 4 generated by a, then f(aaaaa ) 5 (aaaaa)G 5 a.] Clearly, f is well defined, for inserting or deleting expressions of the form xx21 or x21x in elements of W(S) corresponds to inserting or deleting the identity in G. To check that f is operation-preserving, observe that f(x1x2 . . . xn ) (y1y2 . . . ym ) 5 f(x1x2 . . . xny1y2. . . ym) 5 (x1x2 ? ? ? xny1y2 ? ? ? ym)G 5 (x1x2 ? ? ? xn)G(y1y2 ? ? ? ym)G. Finally, f is onto G because S generates G. The following corollary is an immediate consequence of Theorem 26.2 and the First Isomorphism Theorem for Groups. Corollary Universal Factor Group Property Every group is isomorphic to a factor group of a free group.

Generators and Relations We have now laid the foundation for defining a group by way of generators and relations. Before giving the definition, we will illustrate the basic idea with an example. EXAMPLE 2 Let F be the free group on the set {a, b} and let N be the smallest normal subgroup of F containing the set {a4, b2, (ab)2}. We will show that F/N is isomorphic to D4. We begin by observing that the mapping f from F onto D4, which takes a to R90 and b to H (horizontal reflection), defines a homomorphism whose kernel contains N. Thus, F/Ker f is isomorphic to D4. On the other hand, we claim that the set K 5 {N, aN, a2N, a3N, bN, abN, a2bN, a3bN} of left cosets of N is F/N itself. To see this, notice that every member of F/N can be generated by starting with N and successively multiplying on the left by various combinations of a’s and b’s. So, it suffices to show that K is closed under multiplication on the left by a and b. It is trivial that K is closed under left multiplication by a. For b, we will do only one of the eight cases. The others can be done in a similar fashion. Consider b(aN). Since b2, abab, a 4 [ N and Nb 5 bN, we

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have baN 5 baNb2 5 babNb 5 a21(abab)Nb 5 a21Nb = a21a4Nb 5 a3Nb 5 a3bN. Upon completion of the other cases (Exercise 3), we know that F/N has at most eight elements. At the same time, we know that F/Ker f has exactly eight elements. Since F/Ker f is a factor group of F/N [indeed, F/Ker f < (F/N)/(Ker f/N)], it follows that F/N also has eight elements and F/N 5 F/Ker f < D4. Definition Generators and Relations

Let G be a group generated by some subset A 5 {a1, a2, . . . , an} and let F be the free group on A. Let W 5 {w1, w2, . . . , wt} be a subset of F and let N be the smallest normal subgroup of F containing W. We say that G is given by the generators a1, a2, . . . , an and the relations w1 5 w2 5 ? ? ? 5 wt 5 e if there is an isomorphism from F/N onto G that carries ai N to ai .

The notation for this situation is G 5 a1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 e. As a matter of convenience, we have restricted the number of generators and relations in our definition to be finite. This restriction is not necessary, however. Also, it is often more convenient to write a relation in implicit form. For example, the relation a21b23ab 5 e is often written as ab 5 b3a. In practice, one does not bother writing down the normal subgroup N that contains the relations. Instead, one just manipulates the generators and treats anything in N as the identity, as our notation suggests. Rather than saying that G is given by a1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 e, many authors prefer to say that G has the presentation a1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 e. Notice that a free group is “free” of relations; that is, the equivalence class containing the empty word is the only relation. We mention in passing the fact that a subgroup of a free group is also a free group. Free groups are of fundamental importance in a branch of algebra known as combinatorial group theory. EXAMPLE 3 The discussion in Example 2 can now be summed up by writing D4 5 a, b | a4 5 b2 5 (ab)2 5 e. EXAMPLE 4 The group of integers is the free group on one letter; that is, Z < a| . (This is the only nontrivial Abelian group that is free.)

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The next theorem formalizes the argument used in Example 2 to prove that the group defined there has eight elements. Theorem 26.3 (Dyck, 1882) Let

G 5 a1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 e and let

G 5 a1, a2, . . . , an | w1 5 w2 5 ? ? ? 5 wt 5 wt11 5 ? ? ? 5 wt1k 5 e. Then G is a homomorphic image of G.

PROOF See Exercise 5. In words, Theorem 26.3 says that if you start with generators and relations for a group G and create a group G by imposing additional relations, then G is a homomorphic image of G. Corollary Largest Group Satisfying Defining Relations If K is a group satisfying the defining relations of a finite group G and |K| $ |G|, then K is isomorphic to G.

PROOF See Exercise 5. EXAMPLE 5 QUATERNIONS Consider the group G 5 a, b | a2 5 5 (ab)2. What does G look like? Formally, of course, G is isomorphic to F/N, where F is free on {a, b} and N is the smallest normal subgroup of F containing b22a2 and (ab)22a2. Now, let H 5 b and S 5 {H, aH}. Then, just as in Example 2, it follows that S is closed under multiplication by a and b from the left. So, as in Example 2, we have G 5 H < aH. Thus, we can determine the elements of G once we know exactly how many elements there are in H. (Here again, the three relations come in.) To do this, first observe that b2 5 (ab)2 5 abab implies b 5 aba. Then a2 5 b2 5 (aba)(aba) 5 aba2ba 5 ab4a and therefore b4 5 e. Hence, H has at most four elements, and therefore G has at most eight—namely, e, b, b2, b3, a, ab, ab2, and ab3. It is conceivable, however, that not all of these eight elements are distinct. For example, Z2 % Z2 satisfies the defining relations and has only four elements. Perhaps it is b2

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the largest group satisfying the relations. How can we show that the eight elements listed above are distinct? Well, consider the group G generated by the matrices A5 c

0 1 d 21 0

and

B5 c

0 i , d i 0

where i 5 "21. Direct calculations show that in G the elements e, B, B2, B3, A, AB, AB2, and AB3 are distinct and that G satisfies the relations A2 5 B2 5 (AB)2. So, it follows from the corollary to Dyck’s Theorem that G is isomorphic to G and therefore G has order 8. The next example illustrates why, in Examples 2 and 5, it is necessary to show that the eight elements listed for the group are distinct. EXAMPLE 6 Let G 5 a, b | a3 5 b9 5 e, a21ba 5 b21. Once again, we let H 5 b and observe that G 5 H < aH < a2H. Thus, G 5 {aib j | 0 # i # 2, 0 # j # 8}, and therefore G has at most 27 elements. But this time we will not be able to find some concrete group of order 27 satisfying the same relations that G does, for notice that b21 5 a21ba implies b 5 (a21ba)21 5 a21b21a. Hence, b 5 ebe 5 a23ba3 5 a22(a21ba)a2 5 a22b21a2 5 a21(a21b21a)a 5 a21ba 5 b21. So, the original three relations imply the additional relation b2 5 e. But b2 5 e 5 b9 further implies b 5 e. It follows, then, that G has at most three distinct elements—namely, e, a, and a2. But Z3 satisfies the defining relations with a 5 1 and b 5 0. So, |G| 5 3. We hope Example 6 convinces you of the fact that, once a list of the elements of the group given by a set of generators and relations has been obtained, one must further verify that this list has no duplications. Typically, this is accomplished by exhibiting a specific group that satisfies the given set of generators and relations and that has the same size as the list. Obviously, experience plays a role here. Here is a fun example adapted from [1].

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EXAMPLE 7 Let G be the group with the 26 letters of the alphabet as generators. For relations we take strings A 5 B, where A and B are words in some fixed reference, say [2], and have the same pronunciation but different meanings (such words are called homophones). For example, buy 5 by 5 bye, hour 5 our, lead 5 led, whole 5 hole. From these strings and cancellation, we obtain u 5 e 5 h 5 a 5 w 5 0/ (/0 is the identity string). With these examples in mind, we ask, What is the group given by these generators and relations? Surprisingly, the answer is the infinite cyclic group generated by v. To verify this, one must show that every letter except v is equivalent to 0/ and that there are no two homophones that contain a different number of v’s. The former can easily be done with common words. For example, from inn 5 in, plumb 5 plum, and knot 5 not, we see that n 5 b 5 k 5 0. / From too 5 to we have o 5 0. / That there are no two homophones in [2] that have a different number of v’s can be verified by simply checking all cases. In contrast, the reference Handbook of Homophones by W. C. Townsend (available at the website http://members.peak.org/~jeremy/ dictionaryclassic/chapters/homophones.php) lists felt/veldt as homophones. Of course, including these makes the group trivial.

Classification of Groups of Order up to 15 The next theorem illustrates the utility of the ideas presented in this chapter. Theorem 26.4 Classification of Groups of Order 8 (Cayley, 1859) Up to isomorphism, there are only five groups of order 8: Z8, Z4 % Z2, Z2 % Z2 % Z2, D4, and the quaternions.

PROOF The Fundamental Theorem of Finite Abelian Groups takes care of the Abelian cases. Now, let G be a non-Abelian group of order 8. Also, let G1 5 a, b | a4 5 b2 5 (ab)2 5 e and let G2 5 a, b | a2 5 b2 5 (ab)2. We know from the preceding examples that G1 is isomorphic to D4 and G2 is isomorphic to the quaternions. Thus, it suffices to show that G must satisfy the defining relations for G1 or G2. It follows from Exercise 35 in Chapter 2 and Lagrange’s Theorem that G has an element of order 4; call it a. Then, if b is any element of G not in a, we know that G 5 a < ab 5 {e, a, a2, a3, b, ab, a2b, a3b}.

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26 | Generators and Relations

Consider the element b2 of G. Which of the eight elements of G can it be? Not b, ab, a2b, or a3b, by cancellation. Not a, for b2 commutes with b and a does not. Not a3, for the same reason. Thus, b2 5 e or b2 5 a2. Suppose b2 5 e. Since a is a normal subgroup of G, we know that bab21 [ a. From this and the fact that |bab21| 5 |a|, we then conclude that bab21 5 a or bab21 5 a21. The first relation would mean that G is Abelian, so we know that bab21 5 a21. But then, since b2 5 e, we have (ab)2 5 e, and therefore G satisfies the defining relations for G1. Finally, if b2 5 a2 holds instead of b2 5 e, we can use bab21 5 a21 to conclude that (ab)2 5 a(bab21)b2 5 aa21b2 5 b2, and therefore G satisfies the defining relations for G2. The classification of the groups of order 8, together with our results on groups of order p2, 2p, and pq from Chapter 24, allow us to classify the groups of order up to 15, with the exception of those of order 12. We already know four groups of order 12—namely, Z12, Z6 % Z2, D6, and A4. An argument along the lines of Theorem 26.4 can be given to show that there is only one more group of order 12. This group, called the dicyclic group of order 12 and denoted by Q6, has presentation a, b | a6 5 e, a3 5 b2, b21ab 5 a21. Table 26.1 lists the groups of order at most 15. We use Q4 to denote the quaternions (see Example 5 in this chapter). Table 26.1 Classification of Groups of Order Up to 15 Order 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Abelian Groups Z1 Z2 Z3 Z4, Z2 % Z2 Z5 Z6 Z7 Z8, Z4 % Z2, Z2 % Z2 % Z2 Z9, Z3 % Z3 Z10 Z11 Z12, Z6 % Z2 Z13 Z14 Z15

Non-Abelian Groups

D3 D4, Q4 D5 D6, A4, Q6 D7

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Characterization of Dihedral Groups As another nice application of generators and relations, we will now give a characterization of the dihedral groups that has been known for more than 100 years. For n $ 3, we have used Dn to denote the group of symmetries of a regular n-gon. Imitating Example 2, one can show that Dn < a, b | an 5 b2 5 (ab)2 5 e (see Exercise 9). By analogy, these generators and relations serve to define D1 and D2 also. (These are also called dihedral groups.) Finally, we define the infinite dihedral group D` as a, b | a2 5 b2 5 e. The elements of D` can be listed as e, a, b, ab, ba, (ab)a, (ba)b, (ab)2, (ba)2, (ab)2a, (ba)2b, (ab)3, (ba)3, . . . . Theorem 26.5 Characterization of Dihedral Groups Any group generated by a pair of elements of order 2 is dihedral.

PROOF Let G be a group generated by a pair of distinct elements of order 2, say, a and b. We consider the order of ab. If |ab| 5 `, then G is infinite and satisfies the relations of D`. We will show that G is isomorphic to D`. By Dyck’s Theorem, G is isomorphic to some factor group of D`, say, D`/H. Now, suppose h [ H and h 2 e. Since every element of D` has one of the forms (ab)i, (ba)i, (ab)ia, or (ba)ib, by symmetry, we may assume that h 5 (ab)i or h 5 (ab)ia. If h 5 (ab)i, we will show that D`/H satisfies the relations for Di given in Exercise 9. Since (ab)i is in H, we have H 5 (ab)iH 5 (abH)i, so that (abH)21 5 (abH)i21. But (ab)21H 5 b21a21H 5 baH, and it follows that aHabHaH 5 a2HbHaH 5 eHbaH 5 baH 5 (abH) 21. Thus, D`/H 5 aH, bH 5 aH, abH (see Exercise 7), and D`/H satisfies the defining relations for Di (use Exercise 9 with x 5 aH and y 5 abH). In particular, G is finite—an impossibility. If h 5 (ab)ia, then H 5 (ab)iaH 5 (ab)iHaH,

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and therefore (abH)i 5 (ab)iH 5 (aH)21 5 a21H 5 aH. It follows that aH, bH 5 aH, abH # abH. However, (abH)2i 5 (aH)2 5 a2H 5 H, so that D`/H is again finite. This contradiction forces H 5 {e} and G to be isomorphic to D`. Finally, suppose that |ab| 5 n. Since G 5 a, b 5 a, ab, we can show that G is isomorphic to Dn by proving that b(ab)b 5 (ab)21, which is the same as ba 5 (ab)21 (see Exercise 9). But (ab)21 5 b21a21 5 ba, since a and b have order 2.

Realizing the Dihedral Groups with Mirrors A geometric realization of D` can be obtained by placing two mirrors facing each other in a parallel position, as shown in Figure 26.1. If we let a and b denote reflections in mirrors A and B, respectively, then ab, viewed as the composition of a and b, represents a translation through twice the distance between the two mirrors to the left, and ba is the translation through the same distance to the right. A aba ab

FF

a

e

FF

B b ba

FF

bab baba

FF

Figure 26.1 The group D`—reflections in parallel mirrors

The finite dihedral groups can also be realized with a pair of mirrors. For example, if we place a pair of mirrors facing each other at a 45° angle, we obtain the group D4. Notice that in Figure 26.2, the effect of reflecting an object in mirror A, then mirror B, is a rotation of twice the angle between the two mirrors (that is, 90°).

Special Topics

F

B

F F

F aba

e A a

F

baba = abab

F

bab

b

F

ba

F

448

ab

Figure 26.2 The group D4—reflections in mirrors at a 45° angle

In Figure 26.3, we see a portion of the pattern produced by reflections in a pair of mirrors set at a 1° angle. The corresponding group is D180. In general, reflections in a pair of mirrors set at the angle 180°/n correspond to the group Dn. As n becomes larger and larger, the mirrors approach a parallel position. In the limiting case, we have the group D`. Β

Α ab

a

e

b ba

bab ba ba

FF FF FF FF

aba

Figure 26.3 The group D180—reflections in mirrors at a 1° angle

We conclude this chapter by commenting on the advantages and disadvantages of using generators and relations to define groups. The principal advantage is that in many situations—particularly in knot theory, algebraic topology, and geometry—groups defined by way of generators and relations arise in a natural way. Within group theory itself, it is often convenient to construct examples and counterexamples with generators and relations. Among the disadvantages of defining a group by generators and relations is the fact that it is often difficult to decide whether or not the group is finite, or even whether or not a particular element is the identity. Furthermore, the same group can be defined with entirely different sets of generators and relations, and, given two groups defined by generators and relations, it is often extremely difficult to decide whether or not these two groups are isomorphic. Nowadays, these questions are frequently tackled with the aid of a computer.

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26 | Generators and Relations

Exercises It don’t come easy. Title of a Song by RINGO STARR, May 1971

1. Let S be a set of distinct symbols. Show that the relation defined on W(S) in this chapter is an equivalence relation. 2. Let n be an even integer. Prove that Dn /Z(Dn) is isomorphic to Dn/2. 3. Verify that the set K in Example 2 is closed under multiplication on the left by b. 4. Show that a, b | a5 5 b2 5 e, ba 5 a2b is isomorphic to Z2. 5. Prove Theorem 26.3 and its corollary. 6. Let G be the group {61, 6i, 6j, 6k} with multiplication defined as in Exercise 52 in Chapter 9. Show that G is isomorphic to a, b | a2 5 b2 5 (ab)2. (Hence, the name “quaternions.”) 7. In any group, show that a, b 5 a, ab. (This exercise is referred to in the proof of Theorem 26.5.) 8. Let a 5 (12)(34) and b 5 (24). Show that the group generated by a and b is isomorphic to D4. 9. Prove that G 5 x, y | x2 5 yn 5 e, xyx 5 y21 is isomorphic to Dn. (This exercise is referred to in the proof of Theorem 26.5.) 10. What is the minimum number of generators needed for Z2 % Z2 % Z2? Find a set of generators and relations for this group. 11. Suppose that x2 5 y2 5 e and yz 5 zxy. Show that xy 5 yx. 12. Let G 5 a, b | a2 5 b4 5 e, ab 5 b3a. a. Express a3b2abab3 in the form bia j where 0 # i # 1 and 0 # j # 3. b. Express b3abab3a in the form bia j where 0 # i # 1 and 0 # j # 3. 13. Let G 5 a, b | a2 5 b2 5 (ab)2. a. Express b2abab3 in the form bia j. b. Express b3abab3a in the form bia j. 14. Let G be the group defined by the following table. Show that G is isomorphic to Dn.

1 2 3 4 5 6 : 2n

1

2

3

4

5

6

???

2n

1 2 3 4 5 6 : 2n

2 1 4 3 6 5 : 2n 2 1

3 2n 5 2 7 4 : 2n 2 2

4 2n 2 1 6 1 8 3 : 2n 2 3

5 2n 2 2 7 2n 9 2 : 2n 2 4

6 2n 2 3 8 2n 2 1 10 1 : 2n 2 5

??? ??? ??? ??? ??? ??? : ???

2n 3 2 5 4 7 : 1

450

Special Topics

15. Let G 5 x, y | x8 5 y2 5 e, yxyx3 5 e. Show that |G| # 16. Assuming that |G| 5 16, find the center of G and the order of xy. 16. Confirm the classification given in Table 26.1 of all groups of orders 1 to 11. 17. Let G be defined by some set of generators and relations. Show that every factor group of G satisfies the relations defining G. 18. Let G 5 s, t | sts 5 tst. Show that the permutations (23) and (13) satisfy the defining relations of G. Explain why this proves that G is non-Abelian. 19. In D12 5 x, y | x2 5 y12 5 e, xyx 5 y21, prove that the subgroup H 5 x, y3 (which is isomorphic to D4) is not a normal subgroup. 20. Let G 5 x, y | x2n 5 e, xn 5 y2, y21xy 5 x21. Show that Z(G) 5 {e, xn}. Assuming that |G| 5 4n, show that G/Z(G) is isomorphic to Dn. (The group G is called the dicyclic group of order 4n.) 21. Let G 5 a, b | a6 5 b3 5 e, b21ab 5 a3. How many elements does G have? To what familiar group is G isomorphic? 22. Let G 5 x, y | x4 5 y4 5 e, xyxy21 5 e. Show that |G| # 16. Assuming that |G| 5 16, find the center of G and show that G/ y2 is isomorphic to D4. 23. Determine the orders of the elements of D`. 1 a b 24. Let G5•£0 1 c § † a, b, c P Z2 ¶ . Prove that G is isomorphic 0 0 1 to D4. 25. Let G 5 a, b, c, d | ab 5 c, bc 5 d, cd 5 a, da 5 b. Determine |G|. 26. Let G 5 a, b|a2 5 e, b2 5 e, aba 5 bab. To what familiar group is G isomorphic? 27. Let G 5 a, b|a3 5 e, b2 5 e, aba21b21 5 e. To what familiar group is G isomorphic? 28. Given an example of a non-Abelian group that has exactly three elements of finite order. 29. Referring to Example 7 in this chapter, show as many letters as you can that are equivalent to ~. 30. Suppose that a group of order 8 has exactly five elements of order 2. Identify the group.

References 1. J.-F. Mestre, R. Schoof, L. Washington, D. Zagier, “Quotient homophones des groupes libres [Homophonic Quotients of Free Groups],” Experimental Mathematics 2 (1993): 153–155.

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2. H. C. Whitford, A Dictionary of American Homophones and Homographs, New York: Teachers College Press, 1962.

Suggested Readings Alexander H. Fran, Jr. and David Singmaster, Handbook of Cubik Math, Hillside, N. J.: Enslow, 1982. This book is replete with the group-theoretic aspects of the Magic Cube. It uses permutation group theory and generators and relations to discuss the solutions to the cube and related results. The book has numerous challenging exercises stated in group-theoretic terms. Lee Neuwirth, “The Theory of Knots,” Scientific American 240 (1979): 110–124. This article shows how a group can be associated with a knotted string. Mathematically, a knot is just a one-dimensional curve situated in three-dimensional space. The theory of knots—a branch of topology— seeks to classify and analyze the different ways of embedding such a curve. Around the beginning of the 20th century, Henri Poincaré observed that important geometric characteristics of knots could be described in terms of group generators and relations—the so-called knot group. Among other knots, Neuwirth describes the construction of the knot group for the trefoil knot pictured. One set of generators and relations for this group is x, y, z | xy 5 yz, zx 5 yz.

The trefoil knot

David Peifer, “An Introduction to Combinatorial Group Theory and the Word Problem,” Mathematics Magazine 70 (1997): 3–10. This article discusses some fundamental ideas and problems regarding groups given by presentations.

Marshall Hall, Jr. Professor Hall was a mathematician in the broadest sense of the word but with a predilection for group theory, geometry and combinatorics. HANS ZASSENHAUS, Notices of the American Mathematical Society

MARSHALL HALL, JR., was born on September 17, 1910, in St. Louis, Missouri. He demonstrated interest in mathematics at the age of 11 when he constructed a seven-place table of logarithms for the positive integers up to 1000. He completed a B.A. degree in 1932 at Yale. After spending a year at Cambridge University, where he worked with Philip Hall, Harold Davenport, and G. H. Hardy, he returned to Yale for his Ph.D. degree. At the outbreak of World War II, he joined Naval Intelligence and had significant success in deciphering both the Japanese codes and the German Enigma messages. These successes helped to turn the tide of the war. After the war, Hall had faculty appointments at the Ohio State University, Caltech, and Emory University. He died on July 4, 1990. Hall’s highly regarded books on group theory and combinatorial theory are classics. His mathematical legacy includes more than

452

120 research papers on group theory, coding theory, and design theory. His 1943 paper on projective planes ranks among the most cited papers in mathematics. Several fundamental concepts as well as a sporadic simple group are identified with Hall’s name. One of Hall’s most celebrated results is his solution to the “Burnside Problem” for exponent 6—that is, a finitely generated group in which the order of every element divides 6 must be finite. Hall influenced both John Thompson and Michael Aschbacher, two of finite group theory’s greatest contributors. It was Hall who suggested Thompson’s Ph.D. dissertation problem. Hall’s Ph.D. students at Caltech included Donald Knuth and Robert McEliece. To find more information about Hall, visit: http://www–groups.dcs.st–and .ac.uk/~history/

27 Symmetry Groups

I’m not good at math, but I do know that the universe is formed with mathematical principles whether I understand them or not, and I am going to let that guide me. BOB DYLAN, Chronicles, Volume One

Isometries In the early chapters of this book, we briefly discussed symmetry groups. In this chapter and the next, we examine this fundamentally important concept in some detail. It is convenient to begin such a discussion with the definition of an isometry (from the Greek isometros, meaning “equal measure”) in Rn. Definition Isometry

An isometry of n-dimensional space Rn is a function from Rn onto Rn that preserves distance.

In other words, a function T from Rn onto Rn is an isometry if, for every pair of points p and q in Rn, the distance from T( p) to T(q) is the same as the distance from p to q. With this definition, we may now make precise the definition of the symmetry group of an n-dimensional figure. Definition Symmetry Group of a Figure in Rn

Let F be a set of points in Rn. The symmetry group of F in Rn is the set of all isometries of Rn that carry F onto itself. The group operation is function composition.

It is important to realize that the symmetry group of an object depends not only on the object, but also on the space in which we view it. For example, the symmetry group of a line segment in R1 has order 2, the symmetry group of a line segment considered as a set of points in R2 has order 4, and the symmetry group of a line segment viewed as a set of points in R3 has infinite order (see Exercise 9). 453

454

Special Topics

Although we have formulated our definitions for all finite dimensions, our chief interest will be the two-dimensional case. It has been known since 1831 that every isometry of R2 is one of four types: rotation, reflection, translation, and glide-reflection (see [1, p. 46]). Rotation about a point in a plane needs no explanation. A reflection across a line L is that transformation that leaves every point of L fixed and takes every point Q, not on L, to the point Q9 so that L is the perpendicular bisector of the line segment from Q to Q9 (see Figure 27.1). The line L is called the axis of reflection. In an xy-coordinate plane, the transformation (x, y) S (x, 2y) is a reflection across the x-axis, whereas (x, y) S (y, x) is a reflection across the line y 5 x. Some authors call an axis of reflective symmetry L a mirror because L acts like a two-sided mirror; that is, the image of a point Q in a mirror placed on the line L is, in fact, the image of Q under the reflection across the line L. Reflections are called opposite isometries because they reverse orientation. For example, the reflected image of a clockwise spiral is a counterclockwise spiral. Similarly, the reflected image of a right hand is a left hand. (See Figure 27.1.) Q Q' L Axis of reflection

Axis of reflection

Figure 27.1 Reflected images

A translation is simply a function that carries all points the same distance in the same direction. For example, if p and q are points in a plane and T is a translation, then the two directed line segments joining p to T( p) and q to T(q) have the same length and direction. A glide-reflection is the product of a translation and a reflection across the line containing the translation line segment. This line is called the glide-axis. In Figure 27.2, the arrow gives the direction and length of the translation, and is contained in the axis of reflection. A glide-reflection is also an opposite isometry. Successive footprints in wet sand are related by a glide-reflection.

p

T(p)

Figure 27.2 Glide-reflection

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Classification of Finite Plane Symmetry Groups Our first goal in this chapter is to classify all finite plane symmetry groups. As we have seen in earlier chapters, the dihedral group Dn is the plane symmetry group of a regular n-gon. (For convenience, call D2 the plane symmetry group of a nonsquare rectangle and D1 the plane symmetry group of the letter “V.” In particular, D2 < Z2 % Z2 and D1 < Z2.) The cyclic groups Zn are easily seen to be plane symmetry groups also. Figure 27.3 is an illustration of an organism whose plane symmetry group consists of four rotations and is isomorphic to Z4. The surprising fact is that the cyclic groups and dihedral groups are the only finite plane symmetry groups. The famous mathematician Hermann Weyl attributes the following theorem to Leonardo da Vinci (1452–1519).

Figure 27.3 Aurelia Insulinda, an organism whose plane symmetry group is Z4

Theorem 27.1 Finite Symmetry Groups in the Plane The only finite plane symmetry groups are Zn and Dn.

PROOF Let G be a finite plane symmetry group of some figure. We first observe that G cannot contain a translation or a glide-reflection, because in either case G would be infinite. Now observing that the composition of two reflections preserves orientation, we know that such a composition is a translation or rotation. When the two reflections

456

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have parallel axes of reflection, there is no fixed point (see Exercise 10 in the Supplementary Exercises for Chapters 1– 4) so the composition is a translation. Thus, every two reflections in G have reflection axes that intersect in some point. Suppose that f and f 9 are two distinct reflections in G. Then because f f9 preserves orientation, we know that f f9 is a rotation. We use the fact from geometry [2; 366] that a finite group of rotations must have a common center, say P. This means that their axes of reflection must intersect at point P. So, we have shown that all the elements of G have the common fixed point P. For convenience, let us denote a rotation about P of s degrees by Rs. Now, among all rotations in G, let b be the smallest positive angle of rotation. (Such an angle exists, since G is finite and R360 belongs to G.) We claim that every rotation in G is some power of Rb. To see this, suppose that Rs is in G. We may assume 0° , s # 360°. Then, b # s and there is some integer t such that tb # s , (t 1 1)b. But, then Rs2tb 5 Rs(Rb)2t is in G and 0 # s 2 tb , b. Since b represents the smallest positive angle of rotation among the elements of G, we must have s 2 tb 5 0, and therefore, Rs 5 (Rb)t. This verifies the claim. For convenience, let us say that |Rb| 5 n. Now, if G has no reflections, we have proved that G 5 Rb < Zn. If G has at least one reflection, say f, then f, fRb, f(Rb)2, . . . , f(Rb)n21 are also reflections. Furthermore, this is the entire set of reflections of G. For if g is any reflection in G, then fg is a rotation, and so fg 5 (Rb)k for some k. Thus, g 5 f21(Rb)k 5 f(Rb)k. So G 5 {R0, Rb, (Rb)2, . . . , (Rb)n21, f, fRb, a(Rb)2, . . . , f(Rb)n21}, and G is generated by the pair of reflections f and fRb. Hence, by our characterization of the dihedral groups (Theorem 26.5), G is the dihedral group Dn.

Classification of Finite Groups of Rotations in R3 One might think that the set of all possible finite symmetry groups in three dimensions would be much more diverse than is the case for two dimensions. Surprisingly, this is not the case. For example, moving to three dimensions introduces only three new groups of rotations. This observation was first made by the physicist and mineralogist Auguste Bravais in 1849, in his study of possible structures of crystals.

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Theorem 27.2 Finite Groups of Rotations in R3 Up to isomorphism, the finite groups of rotations in R3 are Zn, Dn, A4 , S4 , and A5 .

Theorem 27.2, together with the Orbit-Stabilizer Theorem (Theorem 7.3), makes easy work of determining the group of rotations of an object in R3. EXAMPLE 1 We determine the group G of rotations of the solid in Figure 27.4, which is composed of six congruent squares and eight congruent equilateral triangles. We begin by singling out any one of the squares. Obviously, there are four rotations that map this square to itself, and the designated square can be rotated to the location of any of the other five. So, by the Orbit-Stabilizer Theorem (Theorem 7.3), the rotation group has order 4 ? 6 5 24. By Theorem 27.2, G is one of Z24, D12, and S4. But each of the first two groups has exactly two elements of order 4, whereas G has more than two. So, G is isomorphic to S4.

Figure 27.4

The group of rotations of a tetrahedron (the tetrahedral group) is isomorphic to A4; the group of rotations of a cube or an octahedron (the octahedral group) is isomorphic to S4; the group of rotations of a dodecahedron or an icosahedron (the icosahedral group) is isomorphic to A5. (Coxeter [1, pp. 271–273] specifies which portions of the polyhedra are being permuted in each case.) These five solids are illustrated in Figure 27.5.

458

Special Topics

Figure 27.5 The five regular solids as depicted by Johannes Kepler in Harmonices Mundi, Book II (1619)

Exercises Perhaps the most valuable result of all education is the ability to make yourself do the thing you have to do, when it ought to be done, whether you like it or not. THOMAS HENRY HUXLEY, Technical Education

1. 2. 3. 4.

Show that an isometry of Rn is one-to-one. Show that the translations of Rn form a group. Exhibit a plane figure whose plane symmetry group is Z5. Show that the group of rotations in R3 of a 3-prism (that is, a prism with equilateral ends, as in the following figure) is isomorphic to D3.

5. What is the order of the (entire) symmetry group in R3 of a 3-prism? 6. What is the order of the symmetry group in R3 of a 4-prism (a box with square ends that is not a cube)? 7. What is the order of the symmetry group in R3 of an n-prism?

27 | Symmetry Groups

459

8. Show that the symmetry group in R3 of a box of dimensions 20 3 30 3 40 is isomorphic to Z2 % Z2 % Z2. 9. Describe the symmetry group of a line segment viewed as a. a subset of R1, b. a subset of R2, c. a subset of R3. (This exercise is referred to in this chapter.) 10. (From the “Ask Marilyn” column in Parade Magazine, December 11, 1994.) The letters of the alphabet can be sorted into the following categories: 1. FGJLNPQRSZ 2. BCDEK 3. AMTUVWY 4. HIOX What defines the categories? 11. Exactly how many elements of order 4 does the group in Example 1 have? 12. Why is inversion [that is, f(x, y) 5 (2x, 2y) ] not listed as one of the four kinds of isometries in R2 ? 13. Explain why inversion through a point in R3 cannot be realized by a rotation in R3. 14. Reflection in a line L in R3 is the isometry that takes each point Q to the point Q9 with the property that L is a perpendicular bisector of the line segment joining Q and Q9. Describe a rotation that has this same effect. 15. In R2, a rotation fixes a point; in R3, a rotation fixes a line. In R4, what does a rotation fix? Generalize these observations to Rn. 16. Show that an isometry of a plane preserves angles. 17. Show that an isometry of a plane is completely determined by the image of three noncollinear points. 18. Suppose that an isometry of a plane leaves three noncollinear points fixed. Which isometry is it? 19. Suppose that an isometry of a plane fixes exactly one point. What type of isometry must it be? 20. Suppose that A and B are rotations of 180° about the points a and b, respectively. What is A followed by B? How is the composite motion related to the points a and b?

460

Special Topics

References 1. H. S. M. Coxeter, Introduction to Geometry, 2nd ed., New York: Wiley, 1969. 2. Marvin Jay Greenberg, Euclidean and Non-Euclidean Geometries: Development and History, 3rd ed., New York: W. H. Freeman, 1993.

Suggested Reading† Lorraine Foster, “On the Symmetry Group of the Dodecahedron,” Mathematics Magazine 63 (1990): 106–107. It is shown that the group of rotations of a dodecahedron and the group of rotations of an icosahedron are both A5. J. Rosen, Symmetry Discovered, Cambridge: Cambridge University Press, 1975. This excellent book was written for first- or second-year college students. It includes sections on group theory, spatial symmetry, temporal symmetry, and color symmetry, and chapters on symmetry in nature and the uses of symmetry in science. Andrew Watson, “The Mathematics of Symmetry,” New Scientist, October (1990): 45–50. This article discusses how chemists use group theory to understand molecular structure and how physicists use it to study the fundamental forces and particles.

Suggested Website http://en.wikipedia.org/wiki/Symmetry This website has a wealth of material about symmetry. Included are essays, photos, links, and references.

†See

also the Suggested Reading for Chapter 1.

Frieze Groups 28 and Crystallographic Groups Symmetry, considered as a law of regular composition of structural objects, is similar to harmony. More precisely, symmetry is one of its components, while the other component is dissymmetry. In our opinion the whole esthetics of scientific and artistic creativity lies in the ability to feel this where others fail to perceive it. A. V. SHUBNIKOV AND V. A. KOPTSIK,

Symmetry in Science and Art

The Frieze Groups In this chapter, we discuss an interesting collection of infinite symmetry groups that arise from periodic designs in a plane. There are two types of such groups. The discrete frieze groups are the plane symmetry groups of patterns whose subgroups of translations are isomorphic to Z. These kinds of designs are the ones used for decorative strips and for patterns on jewelry, as illustrated in Figure 28.1. In mathematics, familiar examples include the graphs of y 5 sin x, y 5 tan x, y 5 |sin x|, and |y| 5 sin x. After we analyze the discrete frieze groups, we examine the discrete symmetry groups of plane patterns whose subgroups of translations are isomorphic to Z % Z. In previous chapters, it was our custom to view two isomorphic groups as the same group, since we could not distinguish between them algebraically. In the case of the frieze groups, we will soon see that, although some of them are isomorphic as groups (that is, algebraically the same), geometrically they are quite different. To emphasize this difference, we will treat them separately. In each of the following cases, the given pattern extends infinitely far in both directions. A proof that there are exactly seven types of frieze patterns is given in the appendix to [6].

461

462

Special Topics

Figure 28.1 Frieze patterns

The symmetry group of pattern I (Figure 28.2) consists of translations only. Letting x denote a translation to the right of one unit (that is, the distance between two consecutive R’s), we may write the symmetry group of pattern I as F1 5 {xn | n [ Z}.

R

R

R

Figure 28.2 Pattern I

R

28 | Frieze Groups and Crystallographic Groups

463

The group for pattern II (Figure 28.3), like that of pattern I, is infinitely cyclic. Letting x denote a glide-reflection, we may write the symmetry group of pattern II as F2 5 {xn | n [ Z}.

R

R

R

R

R R

R

Figure 28.3 Pattern II

Notice that the translation subgroup of pattern II is just x2. The symmetry group for pattern III (Figure 28.4) is generated by a translation x and a reflection y across the dashed vertical line. (There are infinitely many axes of reflective symmetry, including those midway between consecutive pairs of opposite-facing R’s. Any one will do.) The entire group (the operation is function composition) is F3 5 {xnym | n [ Z, m 5 0 or 1}.

RR

RR

RR

RR

RR

Figure 28.4 Pattern III

Note that the two elements xy and y have order 2, they generate F3, and their product (xy)y 5 x has infinite order. Thus, by Theorem 26.5, F3 is the infinite dihedral group. A geometric fact about pattern III worth mentioning is that the distance between consecutive pairs of vertical reflection axes is half the length of the smallest translation vector. In pattern IV (Figure 28.5), the symmetry group F4 is generated by a translation x and a rotation y of 180° about a point p midway between consecutive R’s (such a rotation is often called a half-turn). This group, like F3, is also infinite dihedral. (Another rotation point lies between a top and bottom R. As in pattern III, the distance between consecutive points of rotational symmetry is half the length of the smallest translation vector.) Therefore, F4 5 {xnym | n [ Z, m 5 0 or m 5 1}. p

R

R

R

R

Figure 28.5 Pattern IV

R

R

R

R

Special Topics

RR

R

R

R

RR p

R

RR

R

R

Figure 28.6 Pattern V

The symmetry group F5 for pattern V (Figure 28.6) is yet another infinite dihedral group generated by a glide-reflection x and a rotation y of 180° about the point p. Notice that pattern V has vertical reflection symmetry xy. The rotation points are midway between the vertical reflection axes. Thus F5 5 {xnym | n [ Z, m 5 0 or m 5 1}. The symmetry group F6 for pattern VI (Figure 28.7) is generated by a translation x and a horizontal reflection y. The group is F6 5 {xnym | n [ Z, m 5 0 or m 5 1}. Note that, since x and y commute, F6 is not infinite dihedral. In fact, F6 is isomorphic to Z % Z2. Pattern VI is invariant under a glide-reflection also, but in this case the glide-reflection is called trivial, since the axis of the glide-reflection is also an axis of reflection. (Conversely, a glidereflection is nontrivial if its glide-axis is not an axis of reflective symmetry for the pattern.)

R R

R R

R R

R R

Figure 28.7 Pattern VI

The symmetry group F7 of pattern VII (Figure 28.8) is generated by a translation x, a horizontal reflection y, and a vertical reflection z. It is isomorphic to the direct product of the infinite dihedral group and Z2. The product of y and z is a 180° rotation. Therefore, F7 5 {x ny mz k | n [ Z, m 5 0 or m 5 1, k 5 0 or k 5 1}.

RR R

RR R R

R

RR R

RR R R

R

464

Figure 28.8 Pattern VII

The preceding discussion is summarized in Figure 28.9. Figure 28.10 provides an identification algorithm for the frieze patterns. In describing the seven frieze groups, we have not explicitly said how multiplication is done algebraically. However, each group element corresponds to some isometry, so multiplication is the same as function

465

28 | Frieze Groups and Crystallographic Groups

Pattern x21 R I

II

Generators e R

x22 R

R

e R

x21 y e RR

xy x RR

e R

x R

R y

xy

x2 y

x21y e RR

R

xy x2 RR

R

V

yx

x21 R R

e R R

x R R

21

y

xy

x21zx21 RR R

ze RR R

xz x RR R

x21yz x21y

yz y

xyz xy

R

VII

R

x y

R

VI

x = translation

Z

x = glide-reflection

Z

x = translation y = vertical reflection

D

x = translation y = rotation of 180°

D

x = glide-reflection y = rotation of 180°

D

x = translation y = horizontal reflection

Z

x

R

IV

x21 R

R

x2 R

R

III

x21y x21 RR

x2 R

x R

Group isomorphism class

x = translation y = horizontal reflection z = vertical reflection

D

Z2

Z2

Figure 28.9 The seven frieze patterns and their groups of symmetries

composition. Thus, we can always use the geometry to determine the product of any particular string of elements. For example, we know that every element of F7 can be written in the form x ny mz k. So, just for fun, let’s determine the appropriate values for n, m, and k for the element g 5 x21yzxz. We may do this simply by looking at the effect that g has on pattern VII. For convenience, we will pick out a particular R in the pattern and trace the action of g one step at a time. To distinguish this R, we enclose it in a shaded box. Also, we draw the axis of the vertical reflection z as a dashed line segment. See Figure 28.11. Now, comparing the starting position of the shaded R with its final position, we see that x21yzxz 5 x22y. Exercise 7 suggests how one may arrive at the same result through purely algebraic manipulation.

466 Special Topics

Is there a vertical reflection? yes

no

Is there a horizontal reflection?

Is there a horizontal reflection or glide-reflection? no

Is there a half-turn?

Is there a horizontal reflection?

R R R R

no

no

III

no

II

R R

R R

yes

R R

R

I R

R

R

R

yes

V R

R R

R

R

R

R

yes

VI

R R R

R

Is there a half-turn?

IV R R

R R

Figure 28.10 Recognition chart for frieze patterns. Adapted from [6, p. 83].

R R

R

R R R

no

R

VII R

R R R

yes

R

yes

R

R

28 | Frieze Groups and Crystallographic Groups

RR R

RR R

RR R

RR R

RR R

RR R

RR R

RR R

RR R

RR R

RR R

R

R

R

RR R

467

R

z

RR R

R

R

R

RR R R

x

RR R

R

R

R

RR R R

z

RR R

R

R

R

RR R R

y

RR R

R

R

R

RR R R

x⫺1

RR R R

RR R

RR R R

R

RR R R

Figure 28.11

The Crystallographic Groups The seven frieze groups catalog all symmetry groups that leave a design invariant under all multiples of just one translation. However, there are 17 additional kinds of discrete plane symmetry groups that arise from infinitely repeating designs in a plane. These groups are the symmetry groups of plane patterns whose subgroups of translations are isomorphic to Z % Z. Consequently, the patterns are invariant under linear combinations of two linearly independent translations. These 17 groups were first studied by 19th-century crystallographers and are often called the plane crystallographic groups. Another term occasionally used for these groups is wallpaper groups. Our approach to the crystallographic groups will be geometric. It is adapted from the excellent article by Schattschneider [5] and the monograph by Crowe [1]. Our goal is to enable the reader to determine which of the 17 plane symmetry groups corresponds to a given periodic pattern. We begin with some examples.

468

Special Topics

Figure 28.12 Study of Regular Division of the Plane with Fish and Birds, 1938. Escher drawing with symmetry group p1. The arrows are translation vectors.

The simplest of the 17 crystallographic groups contains translations only. In Figure 28.12, we present an illustration of a representative pattern for this group (imagine the pattern repeated to fill the entire plane). The crystallographic notation for it is p1. (This notation is explained in [5].) The symmetry group of the pattern in Figure 28.13 contains translations and glide-reflections. This group has no (nonzero) rotational or reflective symmetry. The crystallographic notation for it is pg. Figure 28.14 has translational symmetry and threefold rotational symmetry (that is, the figure can be rotated 120° about certain points and be brought into coincidence with itself). The notation for this group is p3. Representative patterns for all 17 plane crystallographic groups, together with their notations, are given in Figures 28.15 and 28.16. Figure 28.17 uses a triangle motif to illustrate the 17 classes of symmetry patterns.

28 | Frieze Groups and Crystallographic Groups

469

Figure 28.13 Escher-like tessellation by J. L. Teeters, with symmetry group pg (disregarding shading). The solid arrow is a translation vector. The dashed arrows are glide-reflection vectors.

Figure 28.14 Study of Regular Division of the Plane with Human Figures, 1938. Escher drawing with symmetry p3 (disregarding shading). The inserted arrows are translation vectors.

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Special Topics

Figure 28.15 The plane symmetry groups

All designs in Figures 28.15 and 28.16 except pm, p3, and pg are found in [2]. The designs for p3 and pg are based on elements of Chinese lattice designs found in [2]; the design for pm is based on a weaving pattern from the Sandwich Islands, found in [3].

28 | Frieze Groups and Crystallographic Groups

Figure 28.16 The plane symmetry groups

471

472

Special Topics

Figure 28.17 The 17 plane periodic patterns formed using a triangle motif

28 | Frieze Groups and Crystallographic Groups

473

Identification of Plane Periodic Patterns To decide which of the 17 classes any particular plane periodic pattern belongs to, we may use the flowchart presented in Figure 28.18. This is done by determining the rotational symmetry and whether or not the pattern has reflection symmetry or nontrivial glide-reflection symmetry. These three pieces of information will narrow the list of candidates to at most two. The final test, if necessary, is to determine the locations of the centers of rotation. For example, consider the two patterns in Figure 28.19 generated in a hockey-stick motif. Both patterns have a smallest positive rotational symmetry of 120°; both have reflectional and nontrivial glide-reflectional symmetry. Now, according to Figure 28.18, these patterns must be of type p3m1 or p31m. But notice that the pattern on the left has all its threefold centers of rotation on the reflection axis, whereas in the pattern on the right the points where the three blades meet are not on a reflection axis. Thus, the left pattern is p3m1, and the right pattern is p31m. Table 28.1 (reproduced from [5, p. 443]) can also be used to determine the type of periodic pattern and contains two other features that are often useful. A lattice of points of a pattern is a set of images of any particular point acted on by the translation group of the pattern. A lattice unit of a pattern whose translation subgroup is generated by u and v is a parallelogram formed by a point of the pattern and its image under u, v, and u 1 v. The possible lattices for periodic patterns in a plane, together with lattice units, are shown in Figure 28.20. A generating region (or fundamental region) of a periodic pattern is the smallest portion of the lattice unit whose images under the full symmetry group of the pattern cover the plane. Examples of generating regions for the patterns represented in Figures 28.12, 28.13, and 28.14 are given in Figure 28.21. In Figure 28.21, the portion of the lattice unit with vertical bars is the generating region. The only symmetry pattern in which the lattice unit and the generating region coincide is the p1 pattern illustrated in Figure 28.12. Table 28.1 tells what proportion of the lattice unit constitutes the generating region of each plane periodic pattern. Notice that Table 28.1 reveals that the only possible n-fold rotational symmetries occur when n 5 1, 2, 3, 4, and 6. This fact is commonly called the crystallographic restriction. The first proof of this was given by an Englishman, W. Barlow. The information in Table 28.1 can also be used in reverse to create patterns with a specific symmetry group. The patterns in Figure 28.19 were made in this way.

474

yes none

no

Is there a glide-reflection whose glide-axis is not a reflection axis?

yes

cm

no

pm

yes Is there a glide-reflection? 180°

(five of these) Is there a reflection?

yes no

Are there reflection axes in two directions? Is there a glide-reflection?

What is the smallest angle of rotation?

yes 90°

(three of these) Is there a reflection?

no

Are there reflection axes in four directions?

(three of these) Is there a reflection?

yes no

Are all threefold centers on reflection axes?

pg

no

p1

yes no yes

pgg

no

p2

yes no

p4

120°

Special Topics

(four of these) Is there a reflection?

yes no

Are all rotation centers on reflection axes? pmg p4m p4g

p3m1 p31m

p3 60°

(two of these) Is there a reflection?

yes no

p6m p6

Figure 28.18 Identification flowchart for symmetries of plane periodic patterns

yes

no

pmm

cmm

28 | Frieze Groups and Crystallographic Groups

475

p31m

p3m1

Figure 28.19 Patterns generated in a hockey-stick motif

Rectangular

Parallelogram

Square

Rhombic

Hexagonal (Equilateral triangles)

Figure 28.20 Possible lattices for plane periodic patterns

In sharp contrast to the situation for finite symmetry groups, the transition from two-dimensional crystallographic groups to three-dimensional crystallographic groups introduces a great many more possibilities, since the motif is repeated indefinitely by three independent translations. Indeed, there are 230 three-dimensional crystallographic groups (often called space groups). These were independently determined by Fedorov, Schönflies, and Barlow in the 1890s. David Hilbert, one of the leading mathematicians of the 20th century, focused attention on the crystallographic groups in his

476

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Table 28.1 Identification Chart for Plane Periodic Patternsa

Type

Lattice

Highest Order of Rotation

Reflections

Nontrivial GlideGenerating Reflections Region

p1 p2 pm pg cm pmm pmg

Parallelogram Parallelogram Rectangular Rectangular Rhombic Rectangular Rectangular

1 2 1 1 1 2 2

No No Yes No Yes Yes Yes

No No No Yes Yes No Yes

1 unit 1 2 unit 1 2 unit 1 2 unit 1 2 unit 1 4 unit 1 4 unit

pgg cmm

Rectangular Rhombic

2 2

No Yes

Yes Yes

1 4 1 4

unit unit

p4 p4m

Square Square

4 4

No Yes

No Yes

1 4 1 8

unit unit

p4g

Square

4

Yes

Yes

1 8

unit

p3 p3m1

Hexagonal Hexagonal

3 3

No Yes

No Yes

1 3 1 6

unit unit

p31m

Hexagonal

3

Yes

Yes

1 6

unit

p6 p6m

Hexagonal Hexagonal

6 6

No Yes

No Yes

1 6 unit 1 12 unit

Helpful Distinguishing Properties

Parallel reflection axes Perpendicular reflection axes Fourfold centers on reflection axes Fourfold centers not on reflection axes All threefold centers on reflection axes Not all threefold centers on reflection axes

aA

rotation through an angle of 360°/n is said to have order n. A glide-reflection is nontrivial if its glide-axis is not an axis of reflective symmetry for the pattern.

famous lecture in 1900 at the International Congress of Mathematicians in Paris. One of 23 problems he posed was whether or not the number of crystallographic groups in n dimensions is always finite. This was answered affirmatively by L. Bieberbach in 1910. We mention in passing that in four dimensions, there are 4783 symmetry groups for infinitely repeating patterns. As one might expect, the crystallographic groups are fundamentally important in the study of crystals. In fact, a crystal is defined as a rigid

28 | Frieze Groups and Crystallographic Groups

477

Figure 28.21 A lattice unit and generating region for the patterns in Figures 28.12, 28.13, and 28.14. Generating regions are shaded with bars.

body in which the component particles are arranged in a pattern that repeats in three directions (the repetition is caused by the chemical bonding). A grain of salt and a grain of sugar are two examples of common crystals. In crystalline materials, the motif units are atoms, ions, ionic groups, clusters of ions, or molecules. Perhaps it is fitting to conclude this chapter by recounting two episodes in the history of science in which an understanding of symmetry groups was crucial to a great discovery. In 1912, Max von Laue, a young German physicist, hypothesized that a narrow beam of x-rays directed onto a crystal with a photographic film behind it would be

478

Special Topics

deflected (the technical term is “diffracted”) by the unit cell (made up of atoms or ions) and would show up on the film as spots. (See Figure 1.3.) Shortly thereafter, two British scientists, Sir William Henry Bragg and his 22-year-old son William Lawrence Bragg, who was a student, noted that von Laue’s diffraction spots, together with the known information about crystallographic space groups, could be used to calculate the shape of the internal array of atoms. This discovery marked the birth of modern mineralogy. From the first crystal structures deduced by the Braggs to the present, x-ray diffraction has been the means by which the internal structures of crystals are determined. Von Laue was awarded the Nobel Prize in physics in 1914, and the Braggs were jointly awarded the Nobel Prize in physics in 1915. Our second episode took place in the early 1950s, when a handful of scientists were attempting to learn the structure of the DNA molecule— the basic genetic material. One of these was a graduate student named Francis Crick; another was an x-ray crystallographer, Rosalind Franklin. On one occasion, Crick was shown one of Franklin’s research reports and an x-ray diffraction photograph of DNA. At this point, we let Horace Judson [4, pp. 165–166], our source, continue the story. Crick saw in Franklin’s words and numbers something just as important, indeed eventually just as visualizable. There was drama, too: Crick’s insight began with an extraordinary coincidence. Crystallographers distinguish 230 different space groups, of which the face-centered monoclinic cell with its curious properties of symmetry is only one—though in biological substances a fairly common one. The principal experimental subject of Crick’s dissertation, however, was the x-ray diffraction of the crystals of a protein that was of exactly the same space group as DNA. So Crick saw at once the symmetry that neither Franklin nor Wilkins had comprehended, that Perutz, for that matter, hadn’t noticed, that had escaped the theoretical crystallographer in Wilkins’ lab, Alexander Stokes—namely, that the molecule of DNA, rotated a half turn, came back to congruence with itself. The structure was dyadic, one half matching the other half in reverse.

This was a crucial fact. Shortly thereafter, James Watson and Crick built an accurate model of DNA. In 1962, Watson, Crick, and Maurice Wilkins received the Nobel Prize in medicine and physiology for their discovery. The opinion has been expressed that, had Franklin correctly recognized the symmetry of the DNA molecule, she might have been the one to unravel the mystery and receive the Nobel Prize [4, p. 172].

28 | Frieze Groups and Crystallographic Groups

479

Exercises You can see a lot just by looking. YOGI BERRA

1. 2. 3. 4. 5. 6. 7.

8. 9. 10. 11.

Show that the frieze group F6 is isomorphic to Z % Z2. How many nonisomorphic frieze groups are there? In the frieze group F7, write x2yzxz in the form x ny mz k. In the frieze group F7, write x23zxyz in the form x ny mz k. In the frieze group F7, show that yz 5 zy and xy 5 yx. In the frieze group F7, show that zxz 5 x21. Use the results of Exercises 5 and 6 to do Exercises 3 and 4 through symbol manipulation only (that is, without referring to the pattern). (This exercise is referred to in this chapter.) Prove that in F7 the cyclic subgroup generated by x is a normal subgroup. Quote a previous result that tells why the subgroups x, y and x, z must be normal in F7. Look up the word frieze in an ordinary dictionary. Explain why the frieze groups are appropriately named. Determine which of the seven frieze groups is the symmetry group of each of the following patterns. a.

b.

c.

d.

480

Special Topics

e.

f.

12. Determine the frieze group corresponding to each of the following patterns: a. y 5 sin x, b. y 5 |sin x|, c. |y| 5 sin x, d. y 5 tan x, e. y 5 csc x. 13. Determine the symmetry group of the tessellation of the plane exemplified by the brickwork shown.

14. Determine the plane symmetry group for each of the patterns in Figure 28.17. 15. Determine which of the 17 crystallographic groups is the symmetry group of each of the following patterns. a.

b.

c.

d.

28 | Frieze Groups and Crystallographic Groups

481

16. In the following figure, there is a point labeled 1. Let a be the translation of the plane that carries the point labeled 1 to the point labeled a, and let b be the translation of the plane that carries the point labeled 1 to the point labeled b. The image of 1 under the composition of a and b is labeled ab. In the corresponding fashion, label the remaining points in the figure in the form aib j.

β 1

αβ α

17. The patterns made by automobile tire treads in the snow are frieze patterns. An extensive study of automobile tires revealed that only five of the seven frieze patterns occur. Speculate on which two patterns do not occur and give a possible reason why they do not. 18. Locate a nontrivial glide-reflection axis of symmetry in the cm pattern in Figure 28.16. 19. Determine which of the frieze groups is the symmetry group of each of the following patterns. a. ? ? ? D D V D DV? ? ? b. ? ? ? V V ??? c. ? ? ? L L L L ? ? ? d. ? ? ? V V V V ? ? ? e. ? ? ? N N N N ? ? ? f. ? ? ? H H H H ? ? ? g. ? ? ? L L ? ? ? 20. Locate a nontrivial glide-reflection axis of symmetry in the pattern third from the left in the bottom row in Figure 28.17. L

L

References 1. Donald Crowe, Symmetry, Rigid Motions, and Patterns, Arlington, Va.: COMAP, 1986. 2. Daniel S. Dye, A Grammar of Chinese Lattice, Harvard-Yenching Institute Monograph Series, vol. VI, Cambridge, Mass.: Harvard University Press, 1937. (Reprinted as Chinese Lattice Designs, New York: Dover, 1974.)

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3. Owen Jones, The Grammar of Ornament, New York: Van Nostrand Reinhold, 1972. (Reproduction of the same title, first published in 1856 and reprinted in 1910 and 1928.) 4. Horace Freeland Judson, The Eighth Day of Creation, New York: Simon and Schuster, 1979. 5. D. Schattschneider, “The Plane Symmetry Groups: Their Recognition and Notation,” The American Mathematical Monthly 85 (1978): 439–450. 6. D. K. Washburn and D. W. Crowe, Symmetries of Culture: Theory and Practice of Plane Pattern Analysis, Seattle: University of Washington Press, 1988.

Suggested Readings S. Garfunkel et al., For All Practical Purposes, 7th ed., New York: W. H. Freeman, 2006. This book has a well-written, richly illustrated chapter on symmetry in art and nature. W. G. Jackson, “Symmetry in Automobile Tires and the Left-Right Problem,” Journal of Chemical Education 69 (1992): 624–626. This article uses automobile tires as a tool for introducing and explaining the symmetry terms and concepts important in chemistry. This is a collection of Escher’s periodic drawings together with a mathematical discussion of each one. D. Schattschneider, Visions of Symmetry, New York: Harry Abrams, 2002. A loving, lavish, encyclopedic book on the drawings of M. C. Escher. H. von Baeyer, “Impossible Crystals,” Discover 11(2) (1990): 69–78. This article tells how the discovery of nonperiodic tilings of the plane led to the discovery of quasicrystals. The x-ray diffraction patterns of quasicrystals exhibit fivefold symmetry—something that had been thought to be impossible.

28 | Frieze Groups and Crystallographic Groups

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Suggested Websites http://www.mcescher.com/ This is the official website for the artist M. C. Escher. It features many of his prints and most of his 136 symmetry drawings.

http://britton.disted.camosun.bc.ca/jbsymteslk.htm This spectacular website on symmetry and tessellations has numerous activities and links to many other sites on related topics. It is a wonderful website for K–12 teachers and students.

M. C. Escher I never got a pass mark in math. The funny thing is I seem to latch on to mathematical theories without realizing what is happening. M. C. ESCHER

M. C. ESCHER was born on June 17, 1898, in the Netherlands. His artistic work prior to 1937 was dominated by the representation of visible reality, such as landscapes and buildings. Gradually, he became less and less interested in the visible world and became increasingly absorbed in an inventive approach to space. He studied the abstract space-filling patterns used in the Moorish mosaics in the Alhambra in Spain. He also studied the mathematician George Pólya’s paper on the 17 plane crystallographic groups. Instead of the geometric motifs used by the Moors and Pólya, Escher preferred to use animals, plants, or people in his spacefilling prints. Escher was fond of incorporating various mathematical ideas into his works. Among these are infinity, Möbius bands, stellations,

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deformations, reflections, Platonic solids, spirals, and the hyperbolic plane. Although Escher originals are now quite expensive, it was not until 1951 that he derived a significant portion of his income from his prints. Today, Escher is widely known and appreciated as a graphic artist. His prints have been used to illustrate ideas in hundreds of scientific works. Despite this popularity among scientists, however, Escher has never been held in high esteem in traditional art circles. Escher died on March 27, 1972, in Holland. To find more information about Escher and his art, visit the official website of M. C. Escher: http://www.mcescher.com/

George Pólya Thank you Professor Pólya for all your beautiful contributions to mathematics, to science, to education, and to humanity. A toast from FRANK HARARY on the occasion of Pólya’s 90th birthday

GEORGE PÓLYA was born in Budapest, Hungary, on December 13, 1887. He received a teaching certificate from the University of Budapest in languages before turning to philosophy, mathematics, and physics. In 1912, he was awarded a Ph.D. in mathematics. Horrified by Hitler and World War II, Pólya came to the United States in 1940. After two years at Brown University, he went to Stanford University, where he remained until his death in 1985 at the age of 97. In 1924, Pólya published a paper in a crystallography journal in which he classified the plane symmetry groups and provided a fullpage illustration of the corresponding 17 periodic patterns. B. G. Escher, a geologist, sent a copy of the paper to his artist brother, M. C. Escher, who used Pólya’s black-and-white geometric patterns as a guide for making his own interlocking colored patterns featuring birds, reptiles, and fish.

Pólya contributed to many branches of mathematics, and his collected papers fill four large volumes. Pólya is also famous for his books on problem solving and for his teaching. One of his books has sold more than 1,000,000 copies. The Society for Industrial and Applied Mathematics, the London Mathematical Society, and the Mathematical Association of America have prizes named after Pólya. Pólya taught courses and lectured around the country into his 90s. He never learned to drive a car and took his first plane trip at age 75. He was married for 67 years and had no children. For more information about Pólya, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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John H. Conway “He’s definitely world class, yet he has this kind of childlike enthusiasm.” RONALD GRAHAM Speaking of John H. Conway

JOHN H. CONWAY ranks among the most original and versatile contemporary mathematicians. Conway was born in Liverpool, England, on December 26, 1937, and grew up in a rough neighborhood. As a youngster, he was often beaten up by older boys and did not do well in high school. Nevertheless, his mathematical ability earned him a scholarship to Cambridge University, where he excelled. A pattern that uses repeated shapes to cover a flat surface without gaps or overlaps is called a tiling. In 1975, Oxford physicist Roger Penrose invented an important new way of tiling the plane with two shapes. Unlike patterns whose symmetry group is one of the 17 plane crystallographic groups, Penrose patterns can be neither translated nor rotated to coincide with themselves. Many of the remarkable properties of the Penrose patterns were discovered by Conway. In 1993,

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Conway discovered a new prism that can be used to fill three-dimensional space without gaps or overlaps. Conway has made many significant contributions to number theory, group theory, game theory, knot theory, and combinatorics. Among his most important discoveries are three simple groups, which are now named after him. (Simple groups are the basic building blocks of all groups.) Conway is fascinated by games and puzzles. He invented the game Life and the game Sprouts. Conway has received numerous prestigious honors. In 1987 he joined the faculty at Princeton University, where his title is John von Neumann Distinguished Professor of Mathematics. For more information about Conway, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Symmetry 29 and Counting Let us pause to slake our thirst one last time at symmetry’s bubbling spring. TIMOTHY FERRIS, Coming of Age in the Milky Way

Motivation Permutation groups naturally arise in many situations involving symmetrical designs or arrangements. Consider, for example, the task of coloring the six vertices of a regular hexagon so that three are black 6 and three are white. Figure 29.1 shows the a b 5 20 possibilities. 3

However, if these designs appeared on one side of hexagonal ceramic tiles, it would be nonsensical to count the designs shown in Figure 29.1(a) as different, since all six designs shown there can be obtained from one of them by rotating. (A manufacturer would make only one of the six.) In this case, we say that the designs in Figure 29.1(a) are equivalent under the group of rotations of the hexagon. Similarly, the designs in Figure 29.1(b) are equivalent under the group of rotations, as are the designs in Figure 29.1(c) and those in Figure 29.1(d). And, since no design from Figure 29.1(a)–(d) can be obtained from a design in a different part by rotation, we see that the designs within each part of the figure are equivalent to each other but nonequivalent to any design in another figure. However, the designs in Figure 29.1(b) and (c) are equivalent under the dihedral group D6, since the designs in Figure 29.1(b) can be reflected to yield the designs in Figure 29.1(c). For example, for purposes of arranging three black beads and three white beads to form a necklace, the designs shown in Figure 29.1(b) and (c) would be considered equivalent. In general, we say that two designs (arrangements of beads) A and B are equivalent under a group G of permutations of the arrangements if there is an element f in G such that f(A) 5 B. That is, two designs are equivalent under G if they are in the same orbit of G. It follows, then,

487

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(a)

(c)

(d)

(b)

Figure 29.1

that the number of nonequivalent designs under G is simply the number of orbits of designs under G. (The set being permuted is the set of all possible designs or arrangements.) Notice that the designs in Figure 29.1 divide into four orbits under the group of rotations but only three orbits under the group D6, since the designs in Figure 29.1(b) and (c) form a single orbit under D6. Thus, we could obtain all 20 tile designs from just four tiles, but we could obtain all 20 necklaces from just three of them.

Burnside’s Theorem Although the problems we have just posed are simple enough to solve by observation, more complicated ones require a more sophisticated approach. Such an approach was provided by Georg Frobenius in 1887. Frobenius’s theorem did not become widely known until it appeared in the classic book on group theory by William Burnside in 1911. By an accident of history, Frobenius’s theorem has come to be known as “Burnside’s Theorem.” Before stating this theorem, we recall some notation introduced in Chapter 7 and introduce new

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489

notation. If G is a group of permutations on a set S and i [ S, then stabG(i) 5 {f [ G | f(i) 5 i} and orbG(i) 5 {f(i) | f [ G}. For any set X, we use |X| to denote the number of elements in X. Definition Elements Fixed by f

For any group G of permutations on a set S and any f in G, we let fix(f) 5 {i [ S | f(i) 5 i}. This set is called the elements fixed by f (or more simply, “fix of f”).

Theorem 29.1 (Burnside) If G is a finite group of permutations on a set S, then the number of orbits of elements of S under G is

1 a 0fix(f) 0. 0 G 0 f[G PROOF Let n denote the number of pairs (f, i), with f [ G, i [ S, and f(i) 5 i. We begin by counting these pairs in two ways. First, for each particular f in G, the number of such pairs is exactly |fix(f)|. So, n 5 a 0 fix(f) 0. fPG

(1)

Second, for each particular i in S, observe that |stabG(i)| is exactly the number of pairs (f, i) for which f(i) 5 i. So, n 5 a 0stabG (i) 0. iPS

(2)

It follows from Exercise 33 in Chapter 7 that if s and t are in the same orbit of G, then orbG(s) 5 orbG(t), and thus by the Orbit-Stabilizer Theorem (Theorem 7.3) we have |stabG(s)| 5 |G|/|orbG(s)| 5 |G|/|orbG(t)| 5 |stabG(t)|. So, if we choose s [ S and sum over orbG(s), we have a

t[orbG(s)

0stabG (t) 0 5 0orbG (s) 0 0stabG (s) 0 5 0G 0.

(3)

Finally, by summing over all the elements of G, one orbit at a time, it follows from Equations (1), (2), and (3) that # a 0fix(f) 0 5 a 0 stabG (i) 0 5 0 G 0 (number of orbits) iPS

fPG

and the result follows.

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Applications To illustrate how to apply Burnside’s Theorem, let us return to the ceramic tile and necklace problems. In the case of counting hexagonal tiles with three black vertices and three white vertices, the objects being permutated are the 20 possible designs, whereas the group of permutations is the group of six rotational symmetries of a hexagon. Obviously, the identity fixes all 20 designs. We see from Figure 29.1 that rotations of 60°, 180°, or 300° fix none of the 20 designs. Finally, Figure 29.2 shows fix(f) for the rotations of 120° and 240°. These data are collected in Table 29.1.

Figure 29.2 Tile designs fixed by 120° rotation and 240° rotation Table 29.1 Number of Designs Fixed by Element

Element Identity Rotation of 60° Rotation of 120° Rotation of 180° Rotation of 240° Rotation of 300°

20 0 2 0 2 0

So, applying Burnside’s Theorem, we obtain the number of orbits under the group of rotations as 1 (20 1 0 1 2 1 0 1 2 1 0) 5 4. 6 Now let’s use Burnside’s Theorem to count the number of necklace arrangements consisting of three black beads and three white beads. (For the purposes of analysis, we may arrange the beads in the shape of a regular hexagon.) For this problem, two arrangements are equivalent if they are in the same orbit under D6. Figure 29.3 shows the arrangements fixed

Figure 29.3 Bead arrangements fixed by the reflection across a diagonal

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29 | Symmetry and Counting

Table 29.2

Type of Element Identity Rotation of order 2 (180°) Rotation of order 3 (120° or 240°) Rotation of order 6 (60° or 300°) Reflection across diagonal Reflection across side bisector

Number of Elements of This Type

Number of Arrangements Fixed by Type of Element

1 1 2 2 3 3

20 0 2 0 4 0

by a reflection across a diagonal. Table 29.2 summarizes the information needed to apply Burnside’s Theorem. So, there are 1 12

(1 ? 20 1 1 ? 0 1 2 ? 2 1 2 ? 0 1 3 ? 4 1 3 ? 0) 5 3

nonequivalent ways to string three black beads and three white beads on a necklace. Now that we have gotten our feet wet on a few easy problems, let’s try a more difficult one. Suppose that we have the colors red (R), white (W), and blue (B) that can be used to color the edges of a regular tetrahedron (see Figure 5.1). First, observe that there are 36 5 729 colorings without regard to equivalence. How shall we decide when two colorings of the tetrahedron are nonequivalent? Certainly, if we were to pick up a tetrahedron colored in a certain manner, rotate it, and put it back down, we would think of the tetrahedron as being positioned differently rather than as being colored differently (just as if we picked up a die labeled in the usual way and rolled it, we would not say that the die is now differently labeled). So, our permutation group for this problem is just the group of 12 rotations of the tetrahedron shown in Figure 5.1 and is isomorphic to A4. (The group consists of the identity; eight elements of order 3, each of which fixes one vertex; and three elements of order 2, each of which fixes no vertex.) Every rotation permutes the 729 colorings, and to apply Burnside’s Theorem we must determine the size of fix(f) for each of the 12 rotations of the group. Clearly, the identity fixes all 729 colorings. Next, consider the element (234) of order 3, shown in the bottom row, second from the left in Figure 5.1. Suppose that a specific coloring is fixed by this element

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Table 29.3 Nine Colorings Fixed by (234)

Edge 12 13 14 23 34 24

Colorings R R R R R R

R R R W W W

R R R B B B

W W W W W W

W W W R R R

W W W B B B

B B B B B B

B B B R R R

B B B W W W

(that is, the tetrahedron appears to be colored the same before and after this rotation). Since (234) carries edge 12 to edge 13, edge 13 to edge 14, and edge 14 to edge 12, these three edges must agree in color (edge ij is the edge joining vertex i and vertex j). The same argument shows that the three edges 23, 34, and 42 also must agree in color. So, |fix(234)| 5 32, since there are three choices for each of these two sets of three edges. The nine columns in Table 29.3 show the possible colorings of the two sets of three edges. The analogous analysis applies to the other seven elements of order 3. Now consider the rotation (12)(34) of order 2. (See the second tetrahedron in the top row in Figure 5.1.) Since edges 12 and 34 are fixed, they may be colored in any way and will appear the same after the rotation (12)(34). This yields 3 ? 3 choices for those two edges. Since edge 13 is carried to edge 24, these two edges must agree in color. Similarly, edges 23 and 14 must agree. So, we have three choices for the pair of edges 13 and 24 and three choices for the pair of edges 23 and 14. This means that we have 3 ? 3 ? 3 ? 3 ways to color the tetrahedron that will be equivalent under (12)(34). (Table 29.4 gives the complete list of 81 colorings.) So, |fix((12)(34))| 5 34, and the other two elements of order 2 yield the same results. Now that we have analyzed the three types of group elements, we can apply Burnside’s Theorem. In particular, the number of distinct Table 29.4 81 Colorings Fixed by (12)(34) (X and Y can be any of R, W, and B)

Edge 12 34 13 24 23 14

Colorings X Y R R R R

X Y R R W W

X Y R R B B

X Y W W W W

X Y W W R R

X Y W W B B

X Y B B B B

X Y B B R R

X Y B B W W

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colorings of the edges of a tetrahedron with 3 colors is 1 (1 ? 36 1 8 ? 32 1 3 ? 34) 5 87. 12 Surely it would be a difficult task to solve this problem without Burnside’s Theorem. Just as surely, you are wondering who besides mathematicians are interested in counting problems such as the ones we have discussed. Well, chemists are. Indeed, one set of benzene derivatives can be viewed as six carbon atoms arranged in a hexagon with one of the three radicals NH2, COOH, or OH attached at each carbon atom. See Figure 29.4 for one example. OH C

COOH

COOH

COOH

C

C

C

C C

OH

COOH

Figure 29.4 A benzene derivative

So Burnside’s Theorem enables a chemist to determine the number of benzene molecules (see Exercise 4). Another kind of molecule considered by chemists is visualized as a regular tetrahedron with a carbon atom at the center and any of the four radicals HOCH2 (hydroxymethyl), C2H5 (ethyl), Cl (chlorine), or H (hydrogen) at the four vertices. Again, the number of such molecules can be easily counted using Burnside’s Theorem.

Group Action Our informal approach to counting the number of objects that are considered nonequivalent can be made formal as follows. If G is a group and S is a set of objects, we say that G acts on S if there is a homomorphism g from G to sym(S), the group of all permutations on S. (The homomorphism is sometimes called the group action.) For convenience, we denote the image of g under g as gg. Then two objects x and

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y in S are viewed as equivalent under the action of G if and only if gg(x) 5 y for some g in G. Notice that when g is one-to-one, the elements of G may be regarded as permutations on S. On the other hand, when g is not one-to-one, the elements of G may still be regarded as permutations on S, but there are distinct elements g and h in G such that gg and gh induce the same permutation on S [that is, gg(x) 5 gh(x) for all x in S]. Thus, a group acting on a set is a natural generalization of the permutation group concept. As an example of group action, let S be the two diagonals of a square and let G be D4, the group of symmetries of the square. Then gR0, gR180, gD, gD9 are the identity; gR90, gR270, gH, gV interchange the two diagonals; and the mapping g S gg from D4 to sym(S) is a group homomorphism. As a second example, note that GL(n, F), the group of invertible n 3 n matrices with entries from a field F, acts on the set S of n 3 1 column vectors with entries from F by multiplying the vectors on the left by the matrices. In this case, the mapping g S gg from GL(n, F) to sym(S) is a one-to-one homomorphism. We have used group actions several times in this text without calling them that. The proof of Cayley’s Theorem ( Theorem 6.1) has a group G acting on the elements of G; the proofs of Sylow’s Second Theorem and Third Theorem ( Theorems 24.4 and 24.5) have a group acting on the set of conjugates of a Sylow p-subgroup; and the proof of the Generalized Cayley Theorem ( Theorem 25.3) has G acting on the left cosets of a subgroup H.

Exercises The greater the difficulty, the more glory in surmounting it. EPICURUS

1. Determine the number of ways in which the four corners of a square can be colored with two colors. (It is permissible to use a single color on all four corners.) 2. Determine the number of different necklaces that can be made using 13 white beads and three black beads. 3. Determine the number of ways in which the vertices of an equilateral triangle can be colored with five colors so that at least two colors are used. 4. A benzene molecule can be modeled as six carbon atoms arranged in a regular hexagon in a plane. At each carbon atom, one of three radicals NH2, COOH, or OH can be attached. How many such compounds are possible?

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5. Suppose that in Exercise 4 we permit only NH2 and COOH for the radicals. How many compounds are possible? 6. Determine the number of ways in which the faces of a regular dodecahedron (regular 12-sided solid) can be colored with three colors. 7. Determine the number of ways in which the edges of a square can be colored with six colors so that no color is used on more than one edge. 8. Determine the number of ways in which the edges of a square can be colored with six colors with no restriction placed on the number of times a color can be used. 9. Determine the number of different 11-bead necklaces that can be made using two colors. 10. Determine the number of ways in which the faces of a cube can be colored with three colors. 11. Suppose a cake is cut into 6 identical pieces. How many ways can we color the cake with n colors assuming that each piece receives one color? 12. How many ways can the five points of a five-pointed crown be painted if three colors of paint are available? 13. Let G be a finite group and let sym(G) be the group of all permutations on G. For each g in G, let fg denote the element of sym(G) defined by fg(x) 5 gxg21 for all x in G. Show that G acts on itself under the action g S fg. Give an example in which the mapping g S fg is not one-to-one. 14. Let G be a finite group, let H be a subgroup of G, and let S be the set of left cosets of H in G. For each g in G, let gg denote the element of sym(S) defined by gg(xH) 5 gxH. Show that G acts on S under the action g S gg. 15. For a fixed square, let L1 be the perpendicular bisector of the top and bottom of the square and let L2 be the perpendicular bisector of the left and right sides. Show that D4 acts on {L1, L2} and determine the kernel of the mapping g S gg.

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Suggested Readings Norman Biggs, Discrete Mathematics, Oxford: Clarendon Press, 1989. Chapter 20 of this book presents a more detailed treatment of the subject of symmetry and counting. Doris Schattschneider, “Escher’s Combinational Patterns,” Electronic Journal of Combinatorics, 4(2) (1997): R17. This article discusses a combinatorial problem concerning generating periodic patterns that the artist M. C. Escher posed and solved in an algorithmic way. The problem can also be solved by using Burnside’s Theorem. The article can be downloaded free from the website http://www.combinatorics.org/

William Burnside In one of the most abstract domains of thought, he [Burnside] has systematized and amplified its range so that, there, his work stands as a landmark in the widening expanse of knowledge. Whatever be the estimate of Burnside made by posterity, contemporaries salute him as a Master among the mathematicians of his own generation. A. R. FORSYTH

WILLIAM BURNSIDE was born on July 2, 1852, in London. After graduating from Cambridge University in 1875, Burnside was appointed lecturer at Cambridge, where he stayed until 1885. He then accepted a position at the Royal Naval College at Greenwich and spent the rest of his career in that post. Burnside wrote more than 150 research papers in many fields. He is best remembered, however, for his pioneering work in group theory and his classic book Theory of Groups, which first appeared in 1897. Because of Burnside’s emphasis on the abstract approach, many consider him to be the first pure group theorist. One mark of greatness in a mathematician is the ability to pose important and challenging problems—problems that open up new areas of research for future generations. Here, Burnside excelled. It was he

who first conjectured that a group G of odd order has a series of normal subgroups, G 5 G0 $ G1 $ G2 $ ? ? ? $ Gn 5 {e}, such that Gi /Gi11 is Abelian. This extremely important conjecture was finally proved more than 50 years later by Feit and Thompson in a 255-page paper (see Chapter 25 for more on this). In 1994, Efim Zelmanov received the Fields Medal for his work on a variation of one of Burnside’s conjectures. Burnside was elected a Fellow of the Royal Society and awarded two Royal medals. He served as president of the Council of the London Mathematical Society and received its De Morgan medal. Burnside died on August 21, 1927. To find more information about Burnside, visit: http://www-groups.dcs .st-and.ac.uk/~history/

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Cayley Diagraphs 30 of Groups The important thing in science is not so much to obtain new facts as to discover new ways of thinking about them. SIR WILLIAM LAWRENCE, Beyond Reductionism

Motivation In this chapter, we introduce a graphical representation of a group given by a set of generators and relations. The idea was originated by Cayley in 1878. Although this topic is not usually covered in an abstract algebra book, we include it for five reasons: It provides a method of visualizing a group; it connects two important branches of modern mathematics— groups and graphs; it has many applications to computer science; it gives a review of some of our old friends—cyclic groups, dihedral groups, direct products, and generators and relations; and, most importantly, it is fun! Intuitively, a directed graph (or digraph) is a finite set of points, called vertices, and a set of arrows, called arcs, connecting some of the vertices. Although there is a rich and important general theory of directed graphs with many applications, we are interested only in those that arise from groups.

The Cayley Digraph of a Group Definition Cayley Digraph of a Group

Let G be a finite group and let S be a set of generators for G. We define a digraph Cay(S:G), called the Cayley digraph of G with generating set S, as follows. 1. Each element of G is a vertex of Cay(S:G). 2. For x and y in G, there is an arc from x to y if and only if xs 5 y for some s [ S.

To tell from the digraph which particular generator connects two vertices, Cayley proposed that each generator be assigned a color, and that the arrow joining x to xs be colored with the color assigned to s. He called the 498

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30 | Cayley Diagraphs of Groups

resulting figure the color graph of the group. This terminology is still used occasionally. Rather than use colors to distinguish the different generators, we will use solid arrows, dashed arrows, and dotted arrows. In general, if there is an arc from x to y, there need not be an arc from y to x. An arrow emanating from x and pointing to y indicates that there is an arc from x to y. Following are numerous examples of Cayley digraphs. Note that there are several ways to draw the digraph of a group given by a particular generating set. However, it is not the appearance of the digraph that is relevant but the manner in which the vertices are connected. These connections are uniquely determined by the generating set. Thus, distances between vertices and angles formed by the arcs have no significance. (In the digraphs below, a headless arrow joining two vertices x and y indicates that there is an arc from x to y and an arc from y to x. This occurs when the generating set contains both an element and its inverse. For example, a generator of order 2 is its own inverse.) EXAMPLE 1 Z6 5 1. 1 5

0

0

1 4

2

1

3 4 3

2

5

Cay ({1}: Z 6 )

Cay ({1}: Z 6 )

EXAMPLE 2 Z3 % Z2 5 (1, 0), (0, 1). (0, 1)

(1, 0) (0, 0)

(0, 0)

(0, 1) (0, 1)

(1, 0)

(1, 1)

(2, 0)

(2, 1)

(2, 1)

Cay({(1, 0), (0, 1)}: Z 3 ⊕ Z 2)

(2, 0)

(1, 1) (1, 0)

Cay({(1, 0), (0, 1)}: Z 3 ⊕ Z 2)

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EXAMPLE 3 D4 5 R90, H. H

R90 R0

R0

H

R90H

R90

R180 H

R180

R270 H

R90 H

R90H

R270 H

R180 H

R270

R270

R180

Cay({R90, H}: D4 )

Cay({R90, H}: D 4)

EXAMPLE 4 S3 5 (12), (123). (12)

(123)

(12)

(132)

(1)

(23) (13)

(123)

(23)

(132)

(13)

(12) (1)

Cay({(12), (123)}: S3)

(123) Cay({(12), (123)}: S3)

EXAMPLE 5 S3 5 (12), (13). (12) (1)

(13)

(13) (12)

(132)

(123)

(23)

Cay({(12), (13)}: S 3 )

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30 | Cayley Diagraphs of Groups

EXAMPLE 6 A4 5 (12)(34), (123). (123)

(12)(34) (124)

(14)(23)

(134) (123) (1)

(132)

(12)(34)

(234)

(143)

(243)

(13)(24)

(142) Cay({(12)(34), (123)}: A 4 )

EXAMPLE 7 Q4 5 a, b | a4 5 e, a2 5 b2, b21ab 5 a3. b

a

b

e

ab

a

a 2b

a2

a3b

a3 Cay({a, b}: Q4)

EXAMPLE 8 D` 5 a, b | a2 5 b2 5 e. a bab

ba

b

b e

a

Cay({a, b}: D `)

ab

aba

abab

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The Cayley digraph provides a quick and easy way to determine the value of any product of the generators and their inverses. Consider, for example, the product ab3ab22 from the group given in Example 7. To reduce this to one of the eight elements used to label the vertices, we need only begin at the vertex e and follow the arcs from each vertex to the next as specified in the given product. Of course, b21 means traverse the b arc in reverse. (Observations such as b23 5 b also help.) Tracing the product through, we obtain b. Similarly, one can verify or discover other relations among the generators.

Hamiltonian Circuits and Paths Now that we have these directed graphs, what is it that we care to know about them? One question about directed graphs that has been the object of much research was popularized by the Irish mathematician Sir William Hamilton in 1859, when he invented a puzzle called “Around the World.” His idea was to label the 20 vertices of a regular dodecahedron with the names of famous cities. One solves this puzzle by starting at any particular city (vertex) and traveling “around the world,” moving along the arcs in such a way that each other city is visited exactly once before returning to the original starting point. One solution to this puzzle is given in Figure 30.1, where the vertices are visited in the order indicated. Obviously, this idea can be applied to any digraph; that is, one starts at some vertex and attempts to traverse the digraph by moving along 20 1

2

8 19

15

9

3

7

4

6

10

14

5 12 11 18

13 17

Figure 30.1 Around the World

16

30 | Cayley Diagraphs of Groups

503

arcs in such a way that each vertex is visited exactly once before returning to the starting vertex. (To go from x to y, there must be an arc from x to y.) Such a sequence of arcs is called a Hamiltonian circuit in the digraph. A sequence of arcs that passes through each vertex exactly once without returning to the starting point is called a Hamiltonian path. In the rest of this chapter, we concern ourselves with the existence of Hamiltonian circuits and paths in Cayley digraphs. Figures 30.2 and 30.3 show a Hamiltonian path for the digraph given in Example 2 and a Hamiltonian circuit for the digraph given in Example 7, respectively. Is there a Hamiltonian circuit in Cay({(1, 0), (0, 1)}: Z3 % Z2)? More generally, let us investigate the existence of Hamiltonian circuits in Cay({(1, 0), (0, 1)}: Zm % Zn), where m and n are relatively prime and both are greater than 1. Visualize the Cayley digraph as a rectangular grid coordinatized with Zm % Zn, as (0, 0)

(0, 1)

(1, 0)

(1, 1)

(2, 0)

(2, 1)

Figure 30.2 Hamiltonian Path in Cay({(1, 0), (0, 1)}: Z3 % Z2) from (0, 0) to (2, 1). b

e

ab

a

a 2b

a2

a 3b

a3

Figure 30.3 Hamiltonian Circuit in Cay({a, b}: Q4).

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in Figure 30.4. Suppose there is a Hamiltonian circuit in the digraph and (a, b) is some vertex from which the circuit exits horizontally. (Clearly, such a vertex exists.) Then the circuit must exit (a 2 1, b 1 1) horizontally also, for otherwise the circuit passes through (a, b 1 1) twice—see Figure 30.5. Repeating this argument again and again, we see that the circuit exits horizontally from each of the vertices (a, b), (a 2 1, b 1 1), (a 2 2, b 1 2), . . . , which is just the coset (a, b) 1 (21, 1). But when m and n are relatively prime, (21, 1) is the entire group. Obviously, there cannot be a Hamiltonian circuit consisting entirely of horizontal moves. Let us record what we have just proved.

(0, 0)

(0, 1)

(0, 2)

(0, n –1)

(1, 0)

(1, 1)

(1, 2)

(1, n –1) (0, 1) (1, 0)

(m –1, 0)

(m –1, 1)

(m –1, 2)

(m –1, n –1)

Figure 30.4 Cay({(1, 0), (0, 1)}: Zm % Zn). (a 2 1, b 1 1)

(a, b)

Figure 30.5

Theorem 30.1 A Necessary Condition Cay({(1, 0), (0, 1)}: Zm % Zn ) does not have a Hamiltonian circuit when m and n are relatively prime and greater than 1.

What about when m and n are not relatively prime? In general, the answer is somewhat complicated, but the following special case is easy to prove.

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Theorem 30.2 A Sufficient Condition Cay({(1, 0), (0, 1)}: Zm % Zn ) has a Hamiltonian circuit when n divides m. (0, 0 )

(1, 0 )

(0, 1)

(0, 2 )

(1, 1)

(1, 2 )

First 3 3 3 block

(0, 1)

(2, 0 )

(3, 0)

(2, 1)

(2, 2 )

(3, 1)

(1, 0)

(3, 2 ) Repeat path used in first block

(4, 0)

(5, 0)

(4, 1)

(4, 2 )

(5, 1)

(5, 2 )

Repeat path used in first block kth 3 3 3 block

(3k – 1, 0)

(3k – 1, 1)

(3k – 1, 2)

Figure 30.6 Cay({(1, 0), (0, 1)}: Z3k % Z3)

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PROOF Say, m 5 kn. Then we may think of Zm % Zn as k blocks of size n 3 n. (See Figure 30.6 for an example.) Start at (0, 0) and cover the vertices of the top block as follows. Use the generator (0, 1) to move horizontally across the first row to the end. Then use the generator (1, 0) to move vertically to the point below, and cover the remaining points in the second row by moving horizontally. Keep this process up until the point (n 2 1, 0)—the lower left-hand corner of the first block—has been reached. Next, move vertically to the second block and repeat the process used in the first block. Keep this up until the bottom block is covered. Complete the circuit by moving vertically back to (0, 0). Notice that the circuit given in the proof of Theorem 30.2 is easy to visualize but somewhat cumbersome to describe in words. A much more convenient way to describe a Hamiltonian path or circuit is to specify the starting vertex and the sequence of generators in the order in which they are to be applied. In Example 5, for instance, we may start at (1) and alternate the generators (12) and (13) until we return to (1). In Example 3, we may start at R0 and successively apply R90, R90, R90, H, R90, R90, R90, H. When k is a positive integer and a, b, . . . , c is a sequence of group elements, we use k p (a, b, . . . , c) to denote the concatenation of k copies of the sequence (a, b, . . . , c). Thus, 2 p (R90, R90, R90, H) and 2 p (3 p R90, H) both mean R90, R90, R90, H, R90, R90, R90, H. With this notation, we may conveniently denote the Hamiltonian circuit given in Theorem 30.2 as m p ((n 2 1) p (0, 1), (1, 0)). We leave it as an exercise (Exercise 11) to show that if x1, x2, . . . , xn is a sequence of generators determining a Hamiltonian circuit starting at some vertex, then the same sequence determines a Hamiltonian circuit for any starting vertex. From Theorem 30.1, we know that there are some Cayley digraphs of Abelian groups that do not have any Hamiltonian circuits. But Theorem 30.3 shows that each of these Cayley digraphs does have a Hamiltonian path. There are some Cayley digraphs for non-Abelian groups that do not even have Hamiltonian paths, but we will not discuss them here. Theorem 30.3 Abelian Groups Have Hamiltonian Paths Let G be a finite Abelian group, and let S be any (nonempty†) generating set for G. Then Cay(S:G) has a Hamiltonian path.

S is the empty set, it is customary to define S as the identity group. We prefer to ignore this trivial case.

†If

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PROOF We induct on |S|. If |S| 5 1, say, S 5 {a}, then the digraph is just a circle labeled with e, a, a2, . . . , am21, where |a| 5 m. Obviously, there is a Hamiltonian path for this case. Now assume that |S| . 1. Choose some s [ S. Let T 5 S 2 {s}—that is, T is S with s removed— and set H 5 s1, s2, . . . , sn21 where S 5 {s1, s2, . . . , sn} and s 5 sn. (Notice that H may be equal to G.) Because |T| , |S| and H is a finite Abelian group, the induction hypothesis guarantees that there is a Hamiltonian path (a1, a2, . . . , ak) in Cay(T:H). We will show that (a1, a2, . . . , ak, s, a1, a2, . . . , ak, s, . . . , a1, a2, . . . , ak, s, a1, a2, . . . , ak), where a1, a2, . . . , ak occurs |G|/|H| times and s occurs |G|/|H| 2 1 times, is a Hamiltonian path in Cay(S:G). Because S 5 T < {s} and T generates H, the coset Hs generates the factor group G/H. (Since G is Abelian, this group exists.) Hence, the cosets of H are H, Hs, Hs2, . . . , Hsn, where n 5 |G|/|H| 2 1. Starting from the identity element of G, the path given by (a1, a2, . . . , ak) visits each element of H exactly once [because (a1, a2, . . . , ak) is a Hamiltonian path in Cay(T:H)]. The generator s then moves us to some element of the coset Hs. Starting from there, the path (a1, a2, . . . , ak) visits each element of Hs exactly once. Then, s moves us to the coset Hs2, and we visit each element of this coset exactly once. Continuing this process, we successively move to Hs3, Hs4, . . . , Hsn, visiting each vertex in each of these cosets exactly once. Because each vertex of Cay(S:G) is in exactly one coset Hsi, this implies that we visit each vertex of Cay(S:G) exactly once. Thus we have a Hamiltonian path. We next look at Cayley digraphs with three generators. EXAMPLE 9 Let D3 5 r, f | r 3 5 f 2 5 e, rf 5 fr 2. Then a Hamiltonian circuit in Cay({(r, 0), ( f, 0), (e, 1)}: D3 % Z6) is given in Figure 30.7. (f, 0)

(e, 0)

(f, 1)

(e, 1)

(f, 2)

(e, 2)

(f, 3)

(e, 3)

(f, 4)

(e, 4)

(f, 5)

(e, 5)

(rf, 0)

(r, 5)

(r2f, 0)

(r 2, 5)

Figure 30.7

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Although it is not easy to prove, it is true that Cay({(r, 0), ( f, 0), (e, 1)}: Dn % Zm) has a Hamiltonian circuit for all n and m. (See [3].) Example 10 shows the circuit for this digraph when m is even. EXAMPLE 10 Let Dn 5 r, f | r n 5 f 2 5 e, rf 5 fr21. Then a Hamiltonian circuit in Cay({(r, 0), ( f, 0), (e, 1)}: Dn % Zm) with m even is traced in Figure 30.8. The sequence of generators that traces the circuit is m p [(n 2 1) p (r, 0), ( f, 0), (n 2 1) p (r, 0), (e, 1)]. (f, 0)

(e, 0)

(rf, 0)

(r, 0)

(f, 1)

(e, 1)

(r, 1)

...iterate (r2 f,

0)

(r n – 1f, 0)

(r 2,

(r 2,

0)

(r n –1, 0)

(r n –1f, 1)

1)

(r n–1, 1)

Figure 30.8

Some Applications Cayley graphs are natural models for interconnection networks in computer designs, and Hamiltonicity is an important property in relation to sorting algorithms on such networks. One particular Cayley graph that is used to design and analyze interconnection networks of parallel machines is the symmetric group Sn with the set of all transpositions as the generating set. Hamiltonian paths and circuits in Cayley digraphs arise

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in a variety of group theory contexts. A Hamiltonian path in a Cayley digraph of a group is simply an ordered listing of the group elements without repetition. The vertices of the digraph are the group elements, and the arcs of the path are generators of the group. In 1948, R. A. Rankin used these ideas (although not the terminology) to prove that certain bellringing exercises could not be done by the traditional methods employed by bell ringers. (See [1, Chap. 22] for the group-theoretic aspects of bell ringing.) In 1981, Hamiltonian paths in Cayley digraphs were used in an algorithm for creating computer graphics of Escher-type repeating patterns in the hyperbolic plane [2]. This program can produce repeating hyperbolic patterns in color from among various infinite classes of symmetry groups. The program has now been improved so that the user may choose from many kinds of color symmetry. The 2003 Mathematics Awareness Month poster featured one such image (see http://www. mathaware.org/mam/03/index.html). Two Escher drawings and their computer-drawn counterparts are given in Figures 30.9–30.12. In this chapter, we have shown how one may construct a directed graph from a group. It is also possible to associate a group—called the automorphism group—with every directed graph. In fact, several of the 26 sporadic simple groups were first constructed in this way.

Figure 30.9 M. C. Escher’s Circle Limit I

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Figure 30.10 A computer duplication of the pattern of M. C. Escher’s Circle Limit I [2]. The program used a Hamiltonian path in a Cayley digraph of the underlying symmetry group.

Figure 30.11 M. C. Escher’s Circle Limit IV

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511

Figure 30.12 A computer drawing inspired by the pattern of M. C. Escher’s Circle Limit IV [2]. The program used a Hamiltonian path in a Cayley digraph of the underlying symmetry group.

Exercises A mathematician is a machine for turning coffee into theorems. PAUL ERDÖS

1. Find a Hamiltonian circuit in the digraph given in Example 7 different from the one in Figure 30.3. 2. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Z2). 3. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Zm) where m is even. 4. Write the sequence of generators for each of the circuits found in Exercises 1, 2, and 3. 5. Use the Cayley digraph in Example 7 to evaluate the product a3ba21ba3b21.

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6. Let x and y be two vertices of a Cayley digraph. Explain why two paths from x to y in the digraph yield a group relation. That is, an equation of the form a1a2 . . . am 5 b1b2 . . . bn where the ai’s and bj’s are generators of the Cayley digraph. 7. Use the Cayley digraph in Example 7 to verify the relation aba21b21a21b21 5 a2ba3. 8. Identify the following Cayley digraph of a familiar group.

9. Let D4 5 r, f | r 4 5 e 5 f 2, rf 5 fr21. Verify that 6 p [3 p (r, 0), ( f, 0), 3 p (r, 0), (e, 1)] is a Hamiltonian circuit in Cay({(r, 0), ( f, 0), (e, 1)}: D4 % Z6). 10. Draw a picture of Cay({2, 5}: Z8). 11. If s1, s2, . . . , sn is a sequence of generators that determines a Hamiltonian circuit beginning at some vertex, explain why the same sequence determines a Hamiltonian circuit beginning at any point. (This exercise is referred to in this chapter.) 12. Show that the Cayley digraph given in Example 7 has a Hamiltonian path from e to a. 13. Show that there is no Hamiltonian path in Cay({(1, 0), (0, 1)}: Z3 % Z2) from (0, 0) to (2, 0). 14. Draw Cay({2, 3}: Z6). Is there a Hamiltonian circuit in this digraph? 15. a. Let G be a group of order n generated by a set S. Show that a sequence s1, s2, . . . , sn21 of elements of S is a Hamiltonian path in Cay(S:G) if and only if, for all i and j with 1 # i # j , n, we have si si11 ? ? ? sj 2 e. b. Show that the sequence s1s2 ? ? ? sn is a Hamiltonian circuit if and only if s1s2 ? ? ? sn 5 e, and that whenever 1 # i # j , n, we have si si11 ? ? ? sj 2 e.

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16. Let D4 5 a, b | a2 5 b2 5 (ab)4 5 e. Draw Cay({a, b}: D4). Why is it reasonable to say that this digraph is undirected? 17. Let Dn be as in Example 10. Show that 2 p [(n 2 1) p r, f ] is a Hamiltonian circuit in Cay({r, f}: Dn). 18. Let Q8 5 a, b | a8 5 e, a4 5 b2, b21ab 5 a21. Find a Hamiltonian circuit in Cay({a, b}: Q8). 19. Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q8 % Z5). 20. Prove that the Cayley digraph given in Example 6 does not have a Hamiltonian circuit. Does it have a Hamiltonian path? 21. Find a Hamiltonian circuit in Cay({(R90, 0), (H, 0), (R0, 1)}: D4 % Z3). Does this circuit generalize to the case Dn11 % Zn for all n $ 3? 22. Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q8 % Zm) for all even m. 23. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Z3). 24. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q4 % Zm) for all odd m $ 3. 25. Write the sequence of generators that describes the Hamiltonian circuit in Example 9. 26. Let Dn be as in Example 10. Find a Hamiltonian circuit in Cay({(r, 0), ( f, 0), (e, 1)}: D4 % Z5). 27. 28.

29. 30.

Does your circuit generalize to the case Dn % Zn11 for all n $ 4? Prove that Cay({(0, 1), (1, 1)}: Zm % Zn) has a Hamiltonian circuit for all m and n greater than 1. Suppose that a Hamiltonian circuit exists for Cay({(1, 0), (0, 1)}: Zm % Zn) and that this circuit exits from vertex (a, b) vertically. Show that the circuit exits from every member of the coset (a, b) 1 (1, 21) vertically. Let D2 5 r, f | r2 5 f 2 5 e, rf 5 fr21. Find a Hamiltonian circuit in Cay({(r, 0), ( f, 0), (e, 1)}: D2 % Z3). Let Q8 be as in Exercise 18. Find a Hamiltonian circuit in Cay({(a, 0), (b, 0), (e, 1)}: Q8 % Z3).

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31. In Cay({(1, 0), (0, 1)}: Z4 % Z5) find a sequence of generators that visits exactly one vertex twice and all others exactly once and returns to the starting vertex. 32. In Cay({(1, 0), (0, 1)}: Z4 % Z5) find a sequence of generators that visits exactly two vertices twice and all others exactly once and returns to the starting vertex. 33. Find a Hamiltonian circuit in Cay({(1, 0), (0, 1)}: Z4 % Z6). 34. (Factor Group Lemma) Let S be a generating set for a group G, let N be a cyclic normal subgroup of G, and let S 5 {sN | s [ S}. If (a1N, . . . , ar N) is a Hamiltonian circuit in Cay(S:G/N) and the product a1 ? ? ? ar generates N, prove that |N| p (a1, . . . , ar) is a Hamiltonian circuit in Cay(S:G). 35. A finite group is called Hamiltonian if all of its subgroups are normal. (One non-Abelian example is Q4.) Show that Theorem 30.3 can be generalized to include all Hamiltonian groups.

References 1. F. J. Budden, The Fascination of Groups, Cambridge: Cambridge University Press, 1972. 2. Douglas Dunham, John Lindgren, and David Witte, “Creating Repeating Hyperbolic Patterns,” Computer Graphics 15 (1981): 215–223. 3. David Witte, Gail Letzter, and Joseph A. Gallian, “On Hamiltonian Circuits in Cartesian Products of Cayley Digraphs,” Discrete Mathematics 43 (1983): 297–307.

Suggested Readings Frank Budden, “Cayley Graphs for Some Well-Known Groups,” The Mathematical Gazette 69 (1985): 271–278. This article contains the Cayley graphs of A4, Q4, and S4 using a variety of generators and relations. E. L. Burrows and M. J. Clark, “Pictures of Point Groups,” Journal of Chemical Education 51 (1974): 87–90. Chemistry students may be interested in reading this article. It gives a comprehensive collection of the Cayley digraphs of groups important to chemists.

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Douglas Dunham, John Lindgren, and David Witte, “Creating Repeating Hyperbolic Patterns,” Computer Graphics 15 (1981): 215–223. In this beautifully illustrated paper, a process for creating repeating patterns of the hyperbolic plane is described. The paper is a blend of group theory, geometry, and art. Joseph A. Gallian, “Circuits in Directed Grids,” The Mathematical Intelligencer 13 (1991): 40–43. This article surveys research done on variations of the themes discussed in this chapter. Joseph A. Gallian and David Witte, “Hamiltonian Checkerboards,” Mathematics Magazine 57 (1984): 291–294. http://www.d.umn.edu/~jgallian/checker.pdf This paper gives some additional examples of Hamiltonian circuits in Cayley digraphs. Paul Hoffman, “The Man Who Loves Only Numbers,” The Atlantic Monthly 260 (1987): 60–74. A charming portrait of Paul Erdös, a prolific and eccentric mathematician who contributed greatly to graph theory. Henry Levinson, “Cayley Diagrams,” in Mathematical Vistas: Papers from the Mathematics Section, New York Academy of Sciences, J. Malkevitch and D. McCarthy, eds., 1990: 62–68. This richly illustrated article presents Cayley digraphs of many of the groups that appear in this text. A. T. White, “Ringing the Cosets,” The American Mathematical Monthly 94 (1987): 721–746. This article analyzes the practice of bell ringing by way of Cayley digraphs.

Suggested DVD N is a Number, Mathematical Association of America, 58 minutes. In this documentary, Erdös discusses politics, death, and mathematics. Many of Erdös’s collaborators and friends comment on his work and life. It is available for purchase at www.amazon.com.

Suggested Software Group Explorer is mathematical visualization software that allows users to explore dozens of Cayley digraphs of finite groups visually and interactively. This free software is available at http:// sourceforge.net/projects/groupexplorer

William Rowan Hamilton After Isaac Newton, the greatest mathematician of the English-speaking peoples is William Rowan Hamilton. SIR EDMUND WHITTAKER,

Scientific American

WILLIAM ROWAN HAMILTON was born on August 3, 1805, in Dublin, Ireland. At three, he was skilled at reading and arithmetic. At five, he read and translated Latin, Greek, and Hebrew; at 14, he had mastered 14 languages, including Arabic, Sanskrit, Hindustani, Malay, and Bengali. In 1833, Hamilton provided the first modern treatment of complex numbers. In 1843, he made what he considered his greatest discovery—the algebra of quaternions. The quaternions represent a natural generalization of the complex numbers with three numbers i, j, and k whose squares are 21.

This stamp featuring the quaternions was issued in 1983.

516

With these, rotations in three and four dimensions can be algebraically treated. Of greater significance, however, is the fact that the quaternions are noncommutative under multiplication. This was the first ring to be discovered in which the commutative property does not hold. The essential idea for the quaternions suddenly came to Hamilton after 15 years of fruitless thought! Today Hamilton’s name is attached to several concepts, such as the Hamiltonian function, which represents the total energy in a physical system; the Hamilton-Jacobi differential equations; and the Cayley-Hamilton Theorem from linear algebra. He also coined the terms vector, scalar, and tensor. In his later years, Hamilton was plagued by alcoholism. He died on September 2, 1865, at the age of 60. For more information about Hamilton, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Paul Erdös Paul Erdös is a socially helpless Hungarian who has thought about more mathematical problems than anyone else in history. The Atlantic Monthly

PAUL ERDÖS (pronounced AIR-dish) was one of the best-known and most highly respected mathematicians of the 20th century. Unlike most of his contemporaries, who have concentrated on theory building, Erdös focused on problem solving and problem posing. The problems and methods of solution of Erdös—like those of Euler, whose solutions to special problems pointed the way to much of the mathematical theory we have today—have helped pioneer new theories, such as combinatorial and probabilistic number theory, combinatorial geometry, probabilistic and transfinite combinatorics, and graph theory. Erdös was born on March 26, 1913, in Hungary. Both of his parents were high school mathematics teachers. Erdös, a Jew, left Hungary in 1934 at the age of 21 because of the rapid rise of anti-Semitism in Europe. For the rest of his life he traveled incessantly, rarely pausing more than a month in any one place, giving lectures for

small honoraria and staying with fellow mathematicians. He had little property and no fixed address. All that he owned he carried with him in a medium-sized suitcase, frequently visiting as many as 15 places in a month. His motto was, “Another roof, another proof.” Even in his eighties, he put in 19-hour days doing mathematics. Erdös wrote more than 1500 research papers. He coauthored papers with more than 500 people. These people are said to have Erdös number 1. People who do not have Erdös number 1, but who have written a paper with someone who does, are said to have Erdös number 2, and so on, inductively. Erdös died of a heart attack on September 20, 1996, in Warsaw, Poland. For more information about Erdös, visit: http://www-groups.dcs.st-and .ac.uk/~history/ http://www.oakland.edu/~grossman/ erdoshp.html

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Introduction 31 to Algebraic Coding Theory Damn it, if the machine can detect an error, why can’t it locate the position of the error and correct it? RICHARD W. HAMMING

Motivation One of the most interesting and important applications of finite fields has been the development of algebraic coding theory. This theory, which originated in the late 1940s, was created in response to practical communication problems. (Algebraic coding has nothing to do with secret codes.) Algebraic codes are now used in compact disk and DVD players, fax machines, digital televisions, and bar code scanners, and are essential to computer maintenance. To motivate this theory, imagine that we wish to transmit one of two possible signals to a spacecraft approaching Mars. If the proposed landing site appears unfavorable, we will command the craft to orbit the planet; otherwise, we will command the craft to land. The signal for orbiting will be a 0, and the signal for landing will be a 1. But it is possible that some sort of interference (called noise) could cause an incorrect message to be received. To decrease the chance of this happening, redundancy is built into the transmission process. For example, if we wish the craft to orbit Mars, we could send five 0s. The craft’s onboard computer is programmed to take any five-digit message received and decode the result by majority rule. So, if 00000 is sent and 10001 is received, the computer decides that 0 was the intended message. Notice that, for the computer to make the wrong decision, at least three errors must occur during transmission. If we assume that errors occur independently, it is less likely that three errors will occur than that two or fewer errors will occur. For this reason, this decision process is frequently called the maximum-likelihood decoding procedure. Our particular situation is illustrated in Figure 31.1. The general coding procedure is illustrated in Figure 31.2. 518

519

31 | Introduction to Algebraic Coding Theory

Original message 0 earth

Encoded message 00000 encoder

Received message 10001

Noise transmitter

spacecraft

Decoded message 0

decoder

Figure 31.1 Encoding and decoding by fivefold repetition

Encoded message Original message

encoder

Received message

Noisy channel transmitter

receiver

decoder

Decoded message

Figure 31.2 General encoding-decoding

In practice, the means of transmission are telephone, radiowave, microwave, or even a magnetic disk. The noise might be human error, crosstalk, lightning, thermal noise, or deterioration of a disk. Throughout this chapter, we assume that errors in transmission occur independently. Different methods are needed when this is not the case. Now, let’s consider a more complicated situation. This time, assume that we wish to send a sequence of 0s and 1s of length 500. Further, suppose that the probability that an error will be made in the transmission of any particular digit is .01. If we send this message directly, without any redundancy, the probability that it will be received error free is (.99)500, or approximately .0066. On the other hand, if we adopt a threefold repetition scheme by sending each digit three times and decoding each block of three digits received by majority rule, we can do much better. For example, the sequence 1011 is encoded as 111000111111. If the received message is 011000001110, the decoded message is 1001. Now, what is the probability that our 500-digit message will be error free? Well, if a 1, say, is sent, it will be decoded as a 0 if and only if the block received is 001, 010, 100, or 000. The probability that this will occur is (.01)(.01)(.99) 1 (.01)(.99)(.01) 1 (.99)(.01)(.01) 1 (.01)(.01)(.01) 5 (.01)2[3(.99) 1 .01] 5 .000298 , .0003. Thus, the probability that any particular digit in the sequence will be decoded correctly is greater than .9997, and it follows that the probability that the entire 500-digit message will be decoded correctly is greater than (.9997)500, or approximately .86—a dramatic improvement over .0066.

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This example illustrates the three basic features of a code. There is a set of messages, a method of encoding these messages, and a method of decoding the received messages. The encoding procedure builds some redundancy into the original messages; the decoding procedure corrects or detects certain prescribed errors. Repetition codes have the advantage of simplicity of encoding and decoding, but they are too inefficient. In a fivefold repetition code, 80% of all transmitted information is redundant. The goal of coding theory is to devise message encoding and decoding methods that are reliable, efficient, and reasonably easy to implement. Before plunging into the formal theory, it is instructive to look at a sophisticated example. EXAMPLE 1 The Hamming (7, 4) Code This time, our message set consists of all possible 4-tuples of 0s and 1s (that is, we wish to send a sequence of 0s and 1s of length 4). Encoding will be done by viewing these messages as 1 3 4 matrices with entries from Z2 and multiplying each of the 16 messages on the right by the matrix 1 0 G5≥ 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 1 0

1 0 1 1

0 1 ¥. 1 1

(All arithmetic is done modulo 2.) The resulting 7-tuples are called code words. (See Table 31.1.) Table 31.1 Message

Encoder G

Code Word

0000 0001 0010 0100 1000 1100 1010 1001

→ → → → → → → →

0000000 0001011 0010111 0100101 1000110 1100011 1010001 1001101

Message 0110 0101 0011 1110 1101 1011 0111 1111

Encoder G

Code Word

→ → → → → → → →

0110010 0101110 0011100 1110100 1101000 1011010 0111001 1111111

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Notice that the first four digits of each code word constitute just the original message corresponding to the code word. The last three digits of the code word constitute the redundancy features. For this code, we use the nearest-neighbor decoding method (which, in the case that the errors occur independently, is the same as the maximum-likelihood decoding procedure). For any received word v, we assume that the word sent is the code word v9 that differs from v in the fewest number of positions. If the choice of v9 is not unique, we can decide not to decode or arbitrarily choose one of the code words closest to v. (The first option is usually selected when retransmission is practical.) Once we have decoded the received word, we can obtain the message by deleting the last three digits of v9. For instance, suppose that 1000 were the intended message. It would be encoded and transmitted as u 5 1000110. If the received word were v 5 1100110 (an error in the second position), it would still be decoded as u, since v and u differ in only one position, whereas v and any other code word would differ in at least two positions. Similarly, the intended message 1111 would be encoded as 1111111. If, instead of this, the word 0111111 were received, our decoding procedure would still give us the intended message 1111. The code in Example 1 is one of an infinite class of important codes discovered by Richard Hamming in 1948. The Hamming codes are the most widely used codes. The Hamming (7, 4) encoding scheme can be conveniently illustrated with the use of a Venn diagram, as shown in Figure 31.3. Begin by placing the four message digits in the four overlapping regions I, II, A

I

V

B

A

VI

1

1 0

III IV

II

B

0

VII

1

C

C

0 1

Figure 31.3 Venn diagram of the message 1001 and the encoded message 1001101

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III, and IV, with the digit in position 1 in region I, the digit in position 2 in region II, and so on. For regions V, VI, and VII, assign 0 or 1 so that the total number of 1s in each circle is even. Consider the Venn diagram of the received word 0001101: A

B 0

1

0 0

0 1

1

C

How may we detect and correct an error? Well, observe that each of the circles A and B has an odd number of 1s. This tells us that something is wrong. At the same time, we note that circle C has an even number of 1s. Thus, the portion of the diagram that is in both A and B but not in C is the source of the error. See Figure 31.4. Quite often, codes are used to detect errors rather than correct them. This is especially appropriate when it is easy to retransmit a message. If a received word is not a code word, we have detected an error. For example, computers are designed to use a parity check for numbers. Inside the computer, each number is represented by a string of 0s and 1s. If there is an even number of 1s in this representation, a 0 is attached to the string; if there is an odd number of 1s in the representation, a 1 is attached to the string. Thus, each number stored in the computer memory has an even number of 1s. Now, when the computer reads a A

B

0

1

0

0 1

0 1

C

Figure 31.4 Circles A and B but not C have wrong parity

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number from memory, it performs a parity check. If the read number has an odd number of 1s, the computer will know that an error has been made, and it will reread the number. Note that an even number of errors will not be detected by a parity check. The methods of error detection introduced in Chapters 0 and 5 are based on the same principle. An extra character is appended to a string of numbers so that a particular condition is satisfied. If we find that such a string does not satisfy that condition, we know that an error has occurred.

Linear Codes We now formalize some of the ideas introduced in the preceding discussion. Definition Linear Code

An (n, k) linear code over a finite field F is a k-dimensional subspace V of the vector space

Fn 5 F % F % ? ? ? % F n copies over F. The members of V are called the code words. When F is Z2, the code is called binary.

One should think of an (n, k) linear code over F as a set of n-tuples from F, where each n-tuple has two parts: the message part, consisting of k digits; and the redundancy part, consisting of the remaining n 2 k digits. Note that an (n, k) linear code over a finite field F of order q has q k code words, since every member of the code is uniquely expressible as a linear combination of the k basis vectors with coefficients from F. The set of qk code words is closed under addition and scalar multiplication by members of F. Also, since errors in transmission may occur in any of the n positions, there are qn possible vectors that can be received. Where there is no possibility of confusion, it is customary to denote an n-tuple (a1, a2, . . . , an) more simply as a1a2 ? ? ? an, as we did in Example 1. EXAMPLE 2 The set {0000000, 0010111, 0101011, 1001101, 1100110, 1011010, 0111100, 1110001} is a (7, 3) binary code.

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EXAMPLE 3 The set {0000, 0101, 1010, 1111} is a (4, 2) binary code. Although binary codes are by far the most important ones, other codes are occasionally used. EXAMPLE 4 The set {0000, 0121, 0212, 1022, 1110, 1201, 2011, 2102, 2220} is a (4, 2) linear code over Z3. A linear code over Z3 is called a ternary code. To facilitate our discussion of the error-correcting and errordetecting capability of a code, we introduce the following terminology. Definitions Hamming Distance, Hamming Weight

The Hamming distance between two vectors in Fn is the number of components in which they differ. The Hamming weight of a vector is the number of nonzero components of the vector. The Hamming weight of a linear code is the minimum weight of any nonzero vector in the code.

We will use d(u, v) to denote the Hamming distance between the vectors u and v, and wt(u) for the Hamming weight of the vector u. EXAMPLE 5 Let s 5 0010111, t 5 0101011, u 5 1001101, and v 5 1101101. Then, d(s, t) 5 4, d(s, u) 5 4, d(s, v) 5 5, d(u, v) 5 1; and wt(s) 5 4, wt(t) 5 4, wt(u) 5 4, wt(v) 5 5. The Hamming distance and Hamming weight have the following important properties. Theorem 31.1 Properties of Hamming Distance and Hamming Weight For any vectors u, v, and w, d(u, v) # d(u, w) 1 d(w, v) and d(u, v) 5 wt(u 2 v).

PROOF To prove that d(u, v) 5 wt(u 2 v), simply observe that both d(u, v) and wt(u 2 v) equal the number of positions in which u and v differ. To prove that d(u, v) # d(u, w) 1 d(w, v), note that if u and v differ in the ith position and u and w agree in the ith position, then w and v differ in the ith position.

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With the preceding definitions and Theorem 31.1, we can now explain why the codes given in Examples 1, 2, and 4 will correct any single error, but the code in Example 3 will not. Theorem 31.2 Correcting Capability of a Linear Code If the Hamming weight of a linear code is at least 2t 1 1, then the code can correct any t or fewer errors. Alternatively, the same code can detect any 2t or fewer errors.

PROOF We will use nearest-neighbor decoding; that is, for any received vector v, we will assume that the corresponding code word sent is a code word v9 such that the Hamming distance d(v, v9) is a minimum. (If there is more than one such v9, we do not decode.) Now, suppose that a transmitted code word u is received as the vector v and that at most t errors have been made in transmission. Then, by the definition of distance between u and v, we have d(u, v) # t. If w is any code word other than u, then w 2 u is a nonzero code word. Thus, by assumption, 2t 1 1 # wt(w 2 u) 5 d(w, u) # d(w, v) 1 d(v, u) # d(w, v) 1 t, and it follows that t 1 1 # d(w, v). So, the code word closest to the received vector v is u, and therefore v is correctly decoded as u. To show that the code can detect 2t errors, we suppose that a transmitted code word u is received as the vector v and that at least one error, but no more than 2t errors, was made in transmission. Because only code words are transmitted, an error will be detected whenever a received word is not a code word. But v cannot be a code word, since d(v, u) # 2t, whereas we know that the minimum distance between distinct code words is at least 2t 1 1. Theorem 31.2 is often misinterpreted to mean that a linear code with Hamming weight 2t 1 1 can correct any t errors and detect any 2t or fewer errors simultaneously. This is not the case. The user must choose one or the other role for the code. Consider, for example, the Hamming (7, 4) code given in Table 31.1. By inspection, the Hamming weight of the code is 3 5 2 ? 1 1 1, so we may elect either to correct any single error or to detect any one or two errors. To understand why we can’t do both, consider the received word 0001010. The intended message could have been 0000000, in which case two errors were made (likewise for the intended messages 1011010 and 0101110), or the intended message could have been 0001011, in which case one error was made. But there is no way for us to know which of these possibilities occurred. If our choice were error correction, we would assume—perhaps

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mistakenly—that 0001011 was the intended message. If our choice were error detection, we simply would not decode. (Typically, one would request retransmission.) On the other hand, if we write the Hamming weight of a linear code in the form 2t 1 s 1 1, we can correct any t errors and detect any t 1 s or fewer errors. Thus, for a code with Hamming weight 5, our options include the following: 1. Detect any four errors (t 5 0, s 5 4). 2. Correct any one error and detect any two or three errors (t 5 1, s 5 2). 3. Correct any two errors (t 5 2, s 5 0). EXAMPLE 6 Since the Hamming weight of the linear code given in Example 2 is 4, it will correct any single error and detect any two errors (t 5 1, s 5 1) or detect any three errors (t 5 0, s 5 3). It is natural to wonder how the matrix G used to produce the Hamming code in Example 1 was chosen. Better yet, in general, how can one find a matrix G that carries a subspace V of Fk to a subspace of Fn in such a way that for any k-tuple v in V, the vector vG will agree with v in the first k components and build in some redundancy in the last n 2 k components? Such a matrix is a k 3 n matrix of the form 1 0 ? ? ? 0

0 1 ? ? ? 0

??? ???

???

0 0 ? ? ? 1

a11 ? ? ? ? ak1

???

???

a1n 2 k ? ? ? ? akn 2 k

where the aij’s belong to F. A matrix of this form is called the standard generator matrix (or standard encoding matrix) for the resulting code. Any k 3 n matrix whose rows are linearly independent will transform Fk to a k-dimensional subspace of Fn that could be used to build redundancy, but using the standard generator matrix has the advantage that the original message constitutes the first k components of the transformed vectors. An (n, k) linear code in which the k information digits occur at the beginning of each code word is called a systematic code. Schematically, we have Encoder

message |←k digits→|

message redundant digits |

k

|

n2k

|

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Notice that, by definition, a standard generator matrix produces a systematic code. EXAMPLE 7 From the set of messages {000, 001, 010, 100, 110, 101, 011, 111}, we may construct a (6, 3) linear code over Z2 with the standard generator matrix 1 0 0 1 1 0 G 5 £ 0 1 0 1 0 1 §. 0 0 1 1 1 1 The resulting code words are given in Table 31.2. Since the minimum weight of any nonzero code word is 3, this code will correct any single error or detect any double error. Table 31.2 Message

Encoder G

Code Word

000 001 010 100 110 101 011 111

→ → → → → → → →

000000 001111 010101 100110 110011 101001 011010 111100

EXAMPLE 8 Here we take a set of messages as {00, 01, 02, 10, 11, 12, 20, 21, 22}, and we construct a (4, 2) linear code over Z3 with the standard generator matrix G5c

1 0 2 1 d. 0 1 2 2

The resulting code words are given in Table 31.3. Since the minimum weight of the code is 3, it will correct any single error or detect any double error.

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Table 31.3 Message

Encoder G

Code Word

00 01 02 10 11 12 20 21 22

→ → → → → → → → →

0000 0122 0211 1021 1110 1202 2012 2101 2220

Parity-Check Matrix Decoding Now that we can conveniently encode messages with a standard generator matrix, we need a convenient method for decoding the received messages. Unfortunately, this is not as easy to do; however, in the case where at most one error per code word has occurred, there is a fairly simple method for decoding. (When more than one error occurs in a code word, our decoding method fails.) To describe this method, suppose that V is a systematic linear code over the field F given by the standard generator matrix G 5 [Ik | A], where Ik represents the k 3 k identity matrix and A is the k 3 (n 2 k) matrix obtained from G by deleting the first k columns of G. Then, the n 3 (n 2 k) matrix H5c

2A d, In2k

where 2A is the negative of A and In2k is the (n 2 k) 3 (n 2 k) identity matrix, is called the parity-check matrix for V. (In the literature, the transpose of H is called the parity-check matrix, but H is much more convenient for our purposes.) The decoding procedure is: 1. For any received word w, compute wH. 2. If wH is the zero vector, assume that no error was made. 3. If there is exactly one instance of a nonzero element s [ F and a row i of H such that wH is s times row i, assume that the sent word was w 2 (0 . . . s . . . 0), where s occurs in the ith component. If there is more than one such instance, do not decode. 39. When the code is binary, category 3 reduces to the following. If wH is the ith row of H for exactly one i, assume that an error was

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made in the ith component of w. If wH is more than one row of H, do not decode. 4. If wH does not fit into either category 2 or category 3, we know that at least two errors occurred in transmission and we do not decode. EXAMPLE 9 Consider the Hamming (7, 4) code given in Example 1. The generator matrix is 1 0 G5≥ 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 1 0

1 0 1 1

0 1 ¥, 1 1

and the corresponding parity-check matrix is 1 1 1 H 5 G0 1 0 0

1 0 1 1 0 1 0

0 1 1 1W. 0 0 1

Now, if the received vector is v 5 0000110, we find vH 5 110. Since this is the first row of H and no other row, we assume that an error has been made in the first position of v. Thus, the transmitted code word is assumed to be 1000110, and the corresponding message is assumed to be 1000. Similarly, if w 5 1011111 is the received word, then wH 5 101, and we assume that an error has been made in the second position. So, we assume that 1111111 was sent and that 1111 was the intended message. If the encoded message 1001101 is received as z 5 1001011 (with errors in the fifth and sixth positions), we find zH 5 110. Since this matches the first row of H, we decode z as 0001011 and incorrectly assume that the message 0001 was intended. On the other hand, nearestneighbor decoding would yield the same incorrect result. Notice that when only one error was made in transmission, the parity-check decoding procedure gave us the originally intended message. We will soon see under what conditions this is true, but first we need an important fact relating a code given by a generator matrix and its parity-check matrix.

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Lemma Orthogonality Relation Let C be a systematic (n, k) linear code over F with a standard generator matrix G and parity-check matrix H. Then, for any vector v in F n, we have vH 5 0 (the zero vector) if and only if v belongs to C.

PROOF First note that, since H has rank n 2 k, we may think of H as a linear transformation from F n onto F n2k. Therefore, it follows from the dimension theorem for linear transformations that n 5 n 2 k 1 dim (Ker H), so that Ker H has dimension k. (Alternatively, one can use a group theory argument to show that |Ker H| 5 |F|k.) Then, since the dimension of C is also k, it suffices to show that 2A d . Then, C # Ker H. To do this, let G 5 [Ik | A], so that H 5 c In2k GH 5 [Ik | A] c

2A d 5 2A 1 A 5 [0] In2k

(the zero matrix).

Now, by definition, any vector v in C has the form mG, where m is a message vector. Thus, vH 5 (mG)H 5 m(GH) 5 m[0] 5 0 (the zero vector). Because of the way H was defined, the parity-check matrix method correctly decodes any received word in which no error has been made. But it will do more. Theorem 31.3 Parity-Check Matrix Decoding Parity-check matrix decoding will correct any single error if and only if the rows of the parity-check matrix are nonzero and no one row is a scalar multiple of any other row.

PROOF For simplicity’s sake, we prove only the binary case. In this special situation, the condition on the rows is that they are nonzero and distinct. So, let H be the parity-check matrix, and let’s assume that this condition holds for the rows. Suppose that the transmitted code word w was received with only one error, and that this error occurred in the ith position. Denoting the vector that has a 1 in the ith position and 0s elsewhere by ei, we may write the received word as w 1 ei. Now, using the Orthogonality Lemma, we obtain (w 1 ei)H 5 wH 1 eiH 5 0 1 eiH 5 eiH. But this last vector is precisely the ith row of H. Thus, if there was exactly one error in transmission, we can use the rows of the parity-check

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matrix to identify the location of the error, provided that these rows are distinct. (If two rows, say, the ith and jth, are the same, we know that the error occurred in either the ith position or the jth position, but we do not know in which.) Conversely, suppose that the parity-check matrix method correctly decodes all received words in which at most one error has been made in transmission. If the ith row of the parity-check matrix H were the zero vector and if the code word u 5 0 ? ? ? 0 were received as ei, we would find eiH 5 0 ? ? ? 0, and we would erroneously assume that the vector ei was sent. Thus, no row of H is the zero vector. Now, suppose that the ith row of H and the jth row of H are equal and i 2 j. Then, if some code word w is transmitted and the received word is w 1 ei (that is, there is a single error in the ith position), we find (w 1 ei)H 5 wH 1 eiH 5 ith row of H 5 jth row of H. Thus, our decoding procedure tells us not to decode. This contradicts our assumption that the method correctly decodes all received words in which at most one error has been made.

Coset Decoding There is another convenient decoding method that utilizes the fact that an (n, k) linear code C over a finite field F is a subgroup of the additive group of V 5 Fn. This method was devised by David Slepian in 1956 and is called coset decoding (or standard decoding). To use this method, we proceed by constructing a table, called a standard array. The first row of the table is the set C of code words, beginning in column 1 with the identity 0 ? ? ? 0. To form additional rows of the table, choose an element v of V not listed in the table thus far. Among all the elements of the coset v 1 C, choose one of minimum weight, say, v9. Complete the next row of the table by placing under the column headed by the code word c the vector v9 1 c. Continue this process until all the vectors in V have been listed in the table. [Note that an (n, k) linear code over a field with q elements will have |V:C| 5 qn2k rows.] The words in the first column are called the coset leaders. The decoding procedure is simply to decode any received word w as the code word at the head of the column containing w. EXAMPLE 10 Consider the (6, 3) binary linear code C 5 {000000, 100110, 010101, 001011, 110011, 101101, 011110, 111000}. The first row of a standard array is just the elements of C. Obviously, 100000 is not in C and has minimum weight among the elements of 100000 1 C, so it can be used to lead the second row. Table 31.4 is the completed table.

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Table 31.4 A Standard Array for a (6, 3) Linear Code Words Coset Leaders 000000 100000 010000 001000 000100 000010 000001 100001

100110 000110 110110 101110 100010 100100 100111 000111

010101 110101 000101 011101 010001 010111 010100 110100

001011 101011 011011 000011 001111 001001 001010 101010

110011 010011 100011 111011 110111 110001 110010 010010

101101 001101 111101 100101 101001 101111 101100 001100

011110 111110 001110 010110 011010 011100 011111 111111

111000 011000 101000 110000 111100 111010 111001 011001

If the word 101001 is received, it is decoded as 101101, since 101001 lies in the column headed by 101101. Similarly, the received word 011001 is decoded as 111000. Recall that the first method of decoding that we introduced was the nearest-neighbor method; that is, any received word w is decoded as the code word c such that d(w, c) is a minimum, provided that there is only one code word c such that d(w, c) is a minimum. The next result shows that in this situation, coset decoding is the same as nearestneighbor decoding. Theorem 31.4 Coset Decoding Is Nearest-Neighbor Decoding In coset decoding, a received word w is decoded as a code word c such that d(w, c) is a minimum.

PROOF Let C be a linear code, and let w be any received word. Suppose that v is the coset leader for the coset w 1 C. Then, w 1 C 5 v 1 C, so w 5 v 1 c for some c in C. Thus, using coset decoding, w is decoded as c. Now, if c9 is any code word, then w 2 c9 [ w 1 C 5 v 1 C, so that wt(w 2 c9) $ wt(v), since the coset leader v was chosen as a vector of minimum weight among the members of v 1 C. Therefore, d(w, c9) 5 wt(w 2 c9) $ wt(v) 5 wt(w 2 c) 5 d(w, c). So, using coset decoding, w is decoded as a code word c such that d(w, c) is a minimum. Recall that in our description of nearest-neighbor decoding, we stated that if the choice for the nearest neighbor of a received word v is

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not unique, then we can decide not to decode or to decode v arbitrarily from among those words closest to v. In the case of coset decoding, the decoded value of v is always uniquely determined by the coset leader of the row containing the received word. Of course, this decoded value may not be the word that was sent. When we know a parity-check matrix for a linear code, coset decoding can be considerably simplified. Definition Syndrome

If an (n, k) linear code over F has parity-check matrix H, then, for any vector u in Fn, the vector uH is called the syndrome† of u.

The importance of syndromes stems from the following property. Theorem 31.5 Same Coset—Same Syndrome Let C be an (n, k) linear code over F with a parity-check matrix H. Then, two vectors of Fn are in the same coset of C if and only if they have the same syndrome.

PROOF Two vectors u and v are in the same coset of C if and only if u 2 v is in C. So, by the Orthogonality Lemma, u and v are in the same coset if and only if 0 5 (u 2 v)H 5 uH 2 vH. We may now use syndromes for decoding any received word w: 1. Calculate wH, the syndrome of w. 2. Find the coset leader v such that wH 5 vH. 3. Assume that the vector sent was w 2 v. With this method, we can decode any received word with a table that has only two rows—one row of coset leaders and another row with the corresponding syndromes. EXAMPLE 11 Consider the code given in Example 10. The paritycheck matrix for this code is 1 1 0 H5F 1 0 0 †This

1 0 1 0 1 0

0 1 1 V. 0 0 1

term was coined by D. Hagelbarger in 1959.

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The list of coset leaders and corresponding syndromes is Coset leader

000000 100000

010000 001000

000100 000010 000001 100001

Syndromes

000

101

100

110

011

010

001

111

So, to decode the received word w 5 101001, we compute wH 5 100. Since the coset leader v 5 000100 has 100 as its syndrome, we assume that w 2 000100 5 101101 was sent. If the received word is wr 5 011001, we compute wrH 5 111 and assume w9 2 100001 5 111000 was sent because 100001 is the coset leader with syndrome 111. Notice that these answers are in agreement with those obtained by using the standardarray method of Example 10. The term syndrome is a descriptive term. In medicine, it is used to designate a collection of symptoms that typify a disorder. In coset decoding, the syndrome typifies an error pattern. In this chapter, we have presented algebraic coding theory in its simplest form. A more sophisticated treatment would make substantially greater use of group theory, ring theory, and especially finite-field theory. For example, Gorenstein (see Chapter 25 for a biography) and Zierler, in 1961, made use of the fact that the multiplicative subgroup of a finite field is cyclic. They associated each digit of certain codes with a field element in such a way that an algebraic equation would be derived whose zeros determined the locations of the errors. In some instances, two error-correcting codes are employed. The European Space Agency space probe Giotto, which came within 370 miles of the nucleus of Halley’s Comet in 1986, had two errorcorrecting codes built into its electronics. One code checked for independently occurring errors, and another—a so-called ReedSolomon code—checked for bursts of errors. Giotto achieved an errordetection rate of 0.999999. Reed-Solomon codes are also used on compact discs. They can correct thousands of consecutive errors.

HISTORICAL NOTE

The Ubiquitous Reed-Solomon Codes Irving Reed and Gustave Solomon monitor the encounter of Voyager II with Neptune at the Jet Propulsion Laboratory in 1989.

We conclude this chapter with an adapted version of an article by Barry A. Cipra about the Reed-Solomon codes [1]. It was the first in a series of articles called “Mathematics That Counts” in SIAM News, the news journal of the Society for Industrial and Applied Mathematics. The articles highlight developments in mathematics that have led to products and processes of substantial benefit to industry and the public. In this “Age of Information,” no one need be reminded of the importance not only of speed but also of accuracy in the storage, retrieval, and transmission of data. Machines do make errors, and their non-man-made mistakes can turn otherwise flawless programming into worthless, even dangerous, trash. Just as architects design buildings that will remain standing even through an earthquake, their computer counterparts have come up with sophisticated techniques capable of counteracting digital disasters. The idea for the current error-correcting techniques for everything from computer hard disk drives to CD players was first introduced in 1960 by Irving Reed and Gustave Solomon, then staff members at MIT’s Lincoln Laboratory. . . . “When you talk about CD players and digital audio tape and now digital television, and various other digital imaging systems that are coming—all of those need Reed-Solomon [codes] as an integral part of the system,” says Robert McEliece, a coding theorist in the electrical engineering department at Caltech. Why? Because digital information, virtually by definition, consists of strings of

“bits”—0s and 1s—and a physical device, no matter how capably manufactured, may occasionally confuse the two. Voyager II, for example, was transmitting data at incredibly low power—barely a whisper— over tens of millions of miles. Disk drives pack data so densely that a read/write head can (almost) be excused if it can’t tell where one bit stops and the next 1 (or 0) begins. Careful engineering can reduce the error rate to what may sound like a negligible level—the industry standard for hard disk drives is 1 in 10 billion—but given the volume of information processing done these days, that “negligible” level is an invitation to daily disaster. Error-correcting codes are a kind of safety net—mathematical insurance against the vagaries of an imperfect material world. In 1960, the theory of error-correcting codes was only about a decade old. The basic theory of reliable digital communication had been set forth by Claude Shannon in the late 1940s. At the same time, Richard Hamming introduced an elegant approach to single-error correction and double-error detection. Through the 1950s, a number of 535

researchers began experimenting with a variety of error-correcting codes. But with their SIAM journal paper, McEliece says, Reed and Solomon “hit the jackpot.” The payoff was a coding system based on groups of bits—such as bytes—rather than individual 0s and 1s. That feature makes Reed-Solomon codes particularly good at dealing with “bursts” of errors: six consecutive bit errors, for example, can affect at most two bytes. Thus, even a double-errorcorrection version of a Reed-Solomon code can provide a comfortable safety factor. . . . Mathematically, Reed-Solomon codes are based on the arithmetic of finite fields. Indeed, the 1960 paper begins by defining a code as “a mapping from a vector space of dimension m over a finite field K into a vector space of higher dimension over the same field.” Starting from a “message” (a 0, a1, . . . , am21), where each ak is an element of the field K, a Reed-Solomon code produces (P(0), P(g), P(g2), . . . , P(gN21)), where N is the number of elements in K, g is a generator of the (cyclic) group of nonzero elements in K, and P(x) is the polynomial a0 1 a1x 1 ? ? ? 1 am21x m21. If N is greater than m, then the values of P overdetermine the polynomial, and the properties of finite fields guarantee that the coefficients of P— i.e., the original message—can be recovered from any m of the values . . . . In today’s byte-sized world, for example, it might make sense to let K be the field of order 28, so that each element of K corresponds to a single byte (in computerese, there are four bits to a nibble and two nibbles to a byte). In that case, N 5 28 5 256, and hence messages up to 251 bytes long can be recovered even if two errors occur in transmitting

536

the values P(0), P(g), . . . , P(g255). That’s a lot better than the 1255 bytes required by the say-everything-five-times approach. Despite their advantages, Reed-Solomon codes did not go into use immediately— they had to wait for the hardware technology to catch up. “In 1960, there was no such thing as fast digital electronics”—at least not by today’s standards, says McEliece. The Reed-Solomon paper “suggested some nice ways to process data, but nobody knew if it was practical or not, and in 1960 it probably wasn’t practical.” But technology did catch up, and numerous researchers began to work on implementing the codes. . . . Many other bells and whistles (some of fundamental theoretic significance) have also been added. Compact disks, for example, use a version of a Reed-Solomon code. Reed was among the first to recognize the significance of abstract algebra as the basis for error-correcting codes. “In hindsight it seems obvious,” he told SIAM News. However, he added, “coding theory was not a subject when we published that paper.” The two authors knew they had a nice result; they didn’t know what impact the paper would have. Three decades later, the impact is clear. The vast array of applications, both current and pending, has settled the question of the practicality and significance of ReedSolomon codes. “It’s clear they’re practical, because everybody’s using them now,” says Elwyn Berkekamp. Billions of dollars in modern technology depend on ideas that stem from Reed and Solomon’s original work. In short, says McEliece, “it’s been an extraordinarily influential paper.”

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Exercises The New Testament offers the basis for modern computer coding theory, in the form of an affirmation of the binary number system. “But let your communication be yea, yea; nay, nay: for whatsoever is more than these cometh of evil.” ANONYMOUS

1. Find the Hamming weight of each code word in Table 31.1. 2. Find the Hamming distance between the following pairs of vectors: {1101, 0111}, {0220, 1122}, {11101, 00111}. 3. Referring to Example 1, use the nearest-neighbor method to decode the received words 0000110 and 1110100. 4. For any vector space V and any u, v, w in Fn, prove that the Hamming distance has the following properties: a. d(u, v) 5 d(v, u) (symmetry) b. d(u, v) 5 0 if and only if u 5 v c. d(u, v) 5 d(u 1 w, v 1 w) (translation invariance) 5. Determine the (6, 3) binary linear code with generator matrix 1 0 0 0 1 1 G 5 £ 0 1 0 1 0 1 §. 0 0 1 1 1 0 6. Show that for binary vectors, wt(u 1 v) $ wt(u) 2 wt(v) and equality occurs if and only if for all i the ith component of u is 1 whenever the ith component of v is 1. 7. If the minimum weight of any nonzero code word is 2, what can we say about the error-correcting capability of the code? 8. Suppose that C is a linear code with Hamming weight 3 and that C9 is one with Hamming weight 4. What can C9 do that C can’t? 9. Let C be a binary linear code. Show that the code words of even weight form a subcode of C. (A subcode of a code is a subset of the code that is itself a code.) 10. Let C 5 {0000000, 1110100, 0111010, 0011101, 1001110, 0100111, 1010011, 1101001}. What is the error-correcting capability of C? What is the errordetecting capability of C?

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11. Suppose that the parity-check matrix of a binary linear code is 1 0 H 5 E1 1 0

0 1 1U . 0 1

Can the code correct any single error? 12. Use the generator matrix G5c

1 0 1 1 d 0 1 2 1

to construct a (4, 2) ternary linear code. What is the parity-check matrix for this code? What is the error-correcting capability of this code? What is the error-detecting capability of this code? Use paritycheck decoding to decode the received word 1201. 13. Find all code words of the (7, 4) binary linear code whose generator matrix is 1 0 G5≥ 0 0

0 1 0 0

0 0 1 0

0 0 0 1

1 1 1 0

1 0 1 1

1 1 ¥. 0 1

Find the parity-check matrix of this code. Will this code correct any single error? 14. Show that in a binary linear code, either all the code words end with 0, or exactly half end with 0. What about the other components? 15. Suppose that a code word v is received as the vector u. Show that coset decoding will decode u as the code word v if and only if u 2 v is a coset leader. 16. Consider the binary linear code C 5 {00000, 10011, 01010, 11001, 00101, 10110, 01111, 11100}. Construct a standard array for C. Use nearest-neighbor decoding to decode 11101 and 01100. If the received word 11101 has exactly one error, can we determine the intended code word? If the received word 01100 has exactly one error, can we determine the intended code word?

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17. Construct a (6, 3) binary linear code with generator matrix 1 0 0 1 1 0 G 5 £ 0 1 0 0 1 1 §. 0 0 1 1 0 1 Decode each of the received words 001001, 011000, 000110, 100001 by the following methods: a. nearest-neighbor method, b. parity-check matrix method, c. coset decoding using a standard array, d. coset decoding using the syndrome method. 18. Suppose that the minimum weight of any nonzero code word in a linear code is 6. Discuss the possible options for error correction and error detection. 19. Using the code and the parity-check matrix given in Example 9, show that parity-check matrix decoding cannot detect any multiple errors (that is, two or more errors). 20. Suppose that the last row of a standard array for a binary linear code is 10000 00011 11010 01001 10101 00110 11111 01100. Determine the code. 21. How many code words are there in a (6, 4) ternary linear code? How many possible received words are there for this code? 22. If the parity-check matrix for a binary linear code is 1 0 1 H5F 1 0 0

1 1 0 0 1 0

0 1 1 V, 0 0 1

will the code correct any single error? Why?

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Special Topics

23. Suppose that the parity-check matrix for a ternary code is 2 2 E1 1 0

1 2 2U. 0 1

Can the code correct all single errors? Give a reason for your answer. 24. Prove that for nearest-neighbor decoding, the converse of Theorem 31.2 is true. 25. Can a (6, 3) binary linear code be double-error-correcting using the nearest-neighbor method? Do not assume that the code is systematic. 26. Prove that there is no 2 3 5 standard generator matrix G that will produce a (5, 2) linear code over Z3 capable of detecting all possible triple errors. 27. Why can’t the nearest-neighbor method with a (4, 2) binary linear code correct all single errors? 28. Suppose that one row of a standard array for a binary code is 000100 110000 011110 111101 101010 001001 100111 010011. Determine the row that contains 100001. 29. Use the field F 5 Z2[x]/x2 1 x 1 1 to construct a (5, 2) linear code that will correct any single error. 30. Find the standard generator matrix for a (4, 2) linear code over Z3 that encodes 20 as 2012 and 11 as 1100. Determine the entire code and the parity-check matrix for the code. Will the code correct all single errors? 31. Assume that C is an (n, k) binary linear code and that, for each position i 5 1, 2, . . . , n, the code C has at least one vector with a 1 in the ith position. Show that the average weight of a code word is n/2. 32. Let C be an (n, k) linear code over F such that the minimum weight of any nonzero code word is 2t 1 1. Show that not every vector of weight t 1 1 in Fn can occur as a coset leader. 33. Let C be an (n, k) binary linear code over F 5 Z 2. If v [ Fn but v o C, show that C < (v 1 C) is a linear code. 34. Let C be a binary linear code. Show that either every member of C has even weight or exactly half the members of C have even weight. (Compare with Exercise 19 in Chapter 5.) 35. Let C be an (n, k) linear code. For each i with 1 # i # n, let Ci 5 {v [ C | the ith component of v is 0}. Show that Ci is a subcode of C.

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Reference 1. Barry A. Cipra, “The Ubiquitous Reed-Solomon Codes,” SIAM News 26 (January 1993): 1, 11.

Suggested Readings Norman Levinson, “Coding Theory: A Counterexample to G. H. Hardy’s Conception of Applied Mathematics,” The American Mathematical Monthly 77 (1970): 249–258. The eminent mathematician G. H. Hardy insisted that “real” mathematics was almost wholly useless. In this article, the author argues that coding theory refutes Hardy’s notion. Levinson uses the finite field of order 16 to construct a linear code that can correct any three errors. T. M. Thompson, From Error-Correcting Codes Through Sphere Packing to Simple Groups. Washington, D.C.: The Mathematical Association of America, 1983. Chapter 1 of this award-winning book gives a fascinating historical account of the origins of error-correcting codes.

Richard W. Hamming For introduction of error-correcting codes, pioneering work in operating systems and programming languages, and the advancement of numerical computation. Citation for the Piore Award, 1979

RICHARD W. HAMMING was born in Chicago, Illinois, on February 11, 1915. He graduated from the University of Chicago with a B.S. degree in mathematics. In 1939, he received an M.A. degree in mathematics from the University of Nebraska and, in 1942, a Ph.D. in mathematics from the University of Illinois. During the latter part of World War II, Hamming was at Los Alamos, where he was involved in computing atomic-bomb designs. In 1946, he joined Bell Telephone Laboratories, where he worked in mathematics, computing, engineering, and science. In 1950, Hamming published his famous paper on error-detecting and error-correcting codes. This work started a branch of information theory. The Hamming codes are used in many modern computers. Hamming’s work in the field of numerical analysis has also been of fundamental importance.

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Hamming received numerous prestigious awards, including the Turing Prize from the Association for Computing Machinery, the Piore Award from the Institute of Electrical and Electronics Engineers (IEEE), and the Oender Award from the University of Pennsylvania. In 1986 the IEEE Board of Directors established the Richard W. Hamming Medal “for exceptional contributions to information sciences, systems and technology” and named Hamming as its first recipient. Hamming died of a heart attack on January 7, 1998, at age 82. To find more information about Hamming, visit: http://www-groups.dcs.st-and .ac.uk/~history/

Jessie MacWilliams She was a mathematician who was instrumental in developing the mathematical theory of error-correcting codes from its early development and whose Ph.D. thesis includes one of the most powerful theorems in coding theory. VERA PLESS, SIAM News

An important contributor to coding theory was Jessie MacWilliams. She was born in 1917 in England. After studying at Cambridge University, MacWilliams came to the United States in 1939 to attend Johns Hopkins University. After one year at Johns Hopkins, she went to Harvard for a year. In 1955, MacWilliams became a programmer at Bell Labs, where she learned about coding theory. Although she made a major discovery about codes while a programmer, she could not obtain a promotion to a math research position without a Ph.D. degree. She completed some of the requirements for the Ph.D. while working full-time at Bell Labs and looking after her family. She then returned to Harvard for a year (1961–1962), where she finished her degree. Interestingly, both MacWilliams and her

daughter Ann were studying mathematics at Harvard at the same time. MacWilliams returned to Bell Labs, where she remained until her retirement in 1983. While at Bell Labs, she made many contributions to the subject of error-correcting codes, including The Theory of ErrorCorrecting Codes, written jointly with Neil Sloane. One of her results of great theoretical importance is known as the MacWilliams Identity. She died on May 27, 1990, at the age of 73. To find more information about Mac Williams, visit: http://www.awm-math.org/ noetherbrochure/ MacWilliams80.html

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Vera Pless Vera Pless is a leader in the field of coding theory.

VERA PLESS was born on March 5, 1931, to Russian immigrants on the West Side of Chicago. She accepted a scholarship to attend the University of Chicago at age 15. The program at Chicago emphasized great literature but paid little attention to physics and mathematics. At age 18, with no more than one precalculus course in mathematics, she entered the prestigious graduate program in mathematics at Chicago, where, at that time, there were no women on the mathematics faculty or even women colloquium speakers. After passing her master’s exam, she took a job as a research associate at Northwestern University while pursuing a Ph.D. there. In 1957, she obtained her degree. Over the next several years, Pless stayed at home to raise her children while teaching

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part-time at Boston University. When she decided to work full-time, she found that women were not welcome at most colleges and universities. One person told her outright, “I would never hire a woman.” Fortunately, there was an Air Force Lab in the area that had a group working on error-correcting codes. Although she had never even heard of coding theory, she was hired because of her background in algebra. When the lab discontinued basic research, she took a position as a research associate at MIT in 1972. In 1975, she went to the University of Illinois–Chicago, where she remained until her retirement. During her career, Pless wrote more than 100 research papers, authored a widely used textbook on coding theory, and had 11 Ph. D. students.

An Introduction to 32 Galois Theory Galois theory is a showpiece of mathematical unification, bringing together several different branches of the subject and creating a powerful machine for the study of problems of considerable historical and mathematical importance. IAN STEWART, Galois Theory

Fundamental Theorem of Galois Theory The Fundamental Theorem of Galois Theory is one of the most elegant theorems in mathematics. Look at Figures 32.1 and 32.2. Figure 32.1 depicts the lattice of subgroups of the group of field automorphisms of 4 Q( " 2 , i). The integer along an upward lattice line from a group H1 to a group H2 is the index of H1 in H2. Figure 32.2 shows the lattice of 4 subfields of Q(" 2 , i). The integer along an upward line from a field K1 to a field K2 is the degree of K2 over K1. Notice that the lattice in Figure 32.2 is the lattice of Figure 32.1 turned upside down. This is only one of many relationships between these two lattices. The Fundamental Theorem of Galois Theory relates the lattice of subfields of an algebraic extension E of a field F to the subgroup structure of the group {e, a, a2, a3, b, ab, a2b, a3b} 2

2

{e, a, a 2, a 3}

{e, a 2 , b, a2b } 2

{e, b}

2

2

2

2

2

{e, a 2}

{e, a 2b} 2

{e, a 2 , ab, a 3b }

2

2

2

{e, ab } 2

2

{e, a3b }

2

{ e}

Figure 32.1 Lattice of subgroups of the group of field automorphisms of 4 4 4 4 4 Q(" 2 , i), where a : i S i and " 2 S 2i " 2 , b : i S 2i, and " 2 S " 2 545

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Special Topics 4

Q(√2, i) 2 4

Q(√2) 2

2

2

4

2

2

2

2

Q(√2)

2

4

4

Q(√2, i)

Q(i√2)

Q((11 i) √2)

Q((12 i)√2) 2

2

Q(i√2)

Q(i) 2

2

2

2

Q

Figure 32.2 Lattice of subfields of Q(" 2 , i) 4

of automorphisms of E that send each element of F to itself. This relationship was discovered in the process of attempting to solve a polynomial equation f(x) 5 0 by radicals. Before we can give a precise statement of the Fundamental Theorem of Galois Theory, we need some terminology and notation. Definitions Automorphism, Galois Group, Fixed Field of H

Let E be an extension field of the field F. An automorphism of E is a ring isomorphism from E onto E. The Galois group of E over F, Gal(E/F), is the set of all automorphisms of E that take every element of F to itself. If H is a subgroup of Gal(E/F), the set EH 5 {x [ E | f(x) 5 x for all f [ H} is called the fixed field of H.

It is easy to show that the set of automorphisms of E forms a group under composition. We leave as exercises (Exercises 3 and 5) the verifications that the automorphism group of E fixing F is a subgroup of the automorphism group of E and that, for any subgroup H of Gal(E/F), the fixed field EH of H is a subfield of E. Be careful not to misinterpret Gal(E/F) as something that has to do with factor rings or factor groups. It does not. The following examples will help you assimilate these definitions. In each example, we simply indicate how the automorphisms are defined. We leave to the reader the verifications that the mappings are indeed automorphisms. EXAMPLE 1 Consider the extension Q("2) of Q. Since Q("2) 5 {a 1 b"2 | a, b [ Q}

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547

and any automorphism of a field containing Q must act as the identity on Q (Exercise 1), an automorphism f of Q("2 ) is completely determined by f("2 ). Thus, 2 5 f(2) 5 f("2 "2 ) 5 (f("2 ))2

and therefore f("2) 5 6"2. This proves that the group Gal(Q("2)/Q) has two elements, the identity mapping and the mapping that sends a 1 b"2 to a 2 b "2 . EXAMPLE 2 Consider the extension Q(" 2 ) of Q. An automor3 3 phism f of Q(" 2 ) is completely determined by f(" 2 ). By an argu3 ment analogous to that in Example 1, we see that f(" 2 ) must be a 3 3 cube root of 2. Since Q(" 2 ) is a subset of the real numbers and " 2 3 3 is the only real cube root of 2, we must have f(" 2 ) 5 " 2 . Thus, f 3 is the identity automorphism and Gal(Q(" 2 )/Q) has only one ele3 3 ment. Obviously, the fixed field of Gal(Q(" 2 )/Q) is Q (" 2 ). 3

EXAMPLE 3 Consider the extension Q(" 2 , i) of Q(i). Any auto4 morphism f of Q(" 2 , i) fixing Q(i) is completely determined by 4 f(" 2 ). Since 4

2 5 f(2) 5 f((" 2 )4) 5 (f(" 2 ))4, 4

4

we see that f(" 2 ) must be a fourth root of 2. Thus, there are at most 4 four possible automorphisms of Q(" 2 , i) fixing Q(i). If we define an 4 4 automorphism a such that a(i) 5 i and a(" 2 ) 5 i" 2 , then a [ 4 4 Gal(Q(" 2 , i)/Q(i)) and a has order 4. Thus, Gal(Q(" 2 , i)/Q(i)) is a 2 cyclic group of order 4. The fixed field of {e, a } (where e is the identity 4 automorphism) is Q("2 , i). The lattice of subgroups of Gal(Q(" 2 , i)/ 4 Q(i)) and the lattice of subfields of Q(" 2 , i) containing Q(i) are shown in Figure 32.3. As in Figures 32.1 and 32.2, the integers along the lines 4

4

{e, a, a 2, a 3}

Q(√2, i)

2

2

{e, a 2}

Q(√2, i)

2

2

{e}

Q(i)

Figure 32.3 Lattice of subgroups of Gal(Q(" 2 , i)/Q(i)) 4

and lattice of subfields of Q(" 2 , i) containing Q(i) 4

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Special Topics

of the group lattice represent the index of a subgroup in the group above it, and the integers along the lines of the field lattice represent the degree of the extension of a field over the field below it. EXAMPLE 4 Consider the extension Q("3 , "5 ) of Q. Since

Q("3 , "5 ) 5 {a 1 b"3 1 c"5 1 d"3 "5 | a, b, c, d [ Q},

any automorphism f of Q("3, "5) is completely determined by the two values f("3) and f("5). This time there are four automorphisms: e

a

"3 S "3 "5 S "5

b

"3 S 2"3 "5 S "5

ab

"3 S "3 "5 S 2"5

"3 S 2"3 "5 S 2"5

Obviously, Gal(Q("3, "5)/Q) is isomorphic to Z2 % Z2. The fixed field of {e, a} is Q("5), the fixed field of {e, b} is Q("3), and the fixed field of {e, ab} is Q("3"5). The lattice of subgroups of Gal(Q("3, "5)/Q) and the lattice of subfields of Q("3, "5) are shown in Figure 32.4. {e, a, b, ab}

2

Q(√3,√5 )

{e, b }

{e, a }

2

2

2

2

{e, ab }

2

Q(√5 )

2

2

2

Q (√3 )

2

2

{e}

Q(√3√5 )

2

Q

Figure 32.4 Lattice of subgroups of Gal(Q("3, "5)/Q) and lattice of subfields of Q("3, "5)

Example 5 is a bit more complicated than our previous examples. In particular, the automorphism group is non-Abelian. EXAMPLE 5 Direct calculations show that v 5 21/2 1 i"3/2 satisfies the equations v3 5 1 and v2 1 v 1 1 5 0. Now, consider the 3 extension Q(v, " 2 ) of Q. We may describe the automorphisms of 3 3 Q(v, " 2 ) by specifying how they act on v and " 2 . There are six in all: e

a

b

b2

ab

ab2

vSv v S v2 vSv vSv v S v2 v S v2 3 3 3 3 3 3 3 3 3 3 3 2 2 " 2 S " 2 " 2 S " 2 " 2 S v" 2 " 2 S v " 2 " 2 S v " 2 " 2 S v" 2 3

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32 | An Introduction to Galois Theory

Since ab 2 ba, we know that Gal(Q(v, " 2 )/Q) is isomorphic to S3. (See Theorem 7.2.) The lattices of subgroups and subfields are shown in Figure 32.5. 3

{e, a, b, b 2,ab, ab 2} 2

3

{e, b, b 2}

3

{e, a } 3

3

{e, ab }

2

2

{e, ab 2} 2

{e} 3

Q(ω ,√2) 3 Q(ω )

3

3

3

2 3

Q(ω √2)

Q(√2) 2

2

2

2

3

Q(ω √2) 3

Q

Figure 32.5 Lattice of subgroups of Gal(Q(v, " 2 )/Q) and lattice 3 of subfields of Q(v, " 2 ), where v 5 21/2 1 i"3/2. 3

The lattices in Figure 32.5 have been arranged so that each nontrivial proper field occupying the same position as some group is the fixed field 3 of that group. For instance, Q(v" 2 ) is the fixed field of {e, ab}. The preceding examples show that, in certain cases, there is an intimate connection between the lattice of subfields between E and F and the lattice of subgroups of Gal(E/F). In general, if E is an extension of F, and we let ^ be the lattice of subfields of E containing F and let & be the lattice of subgroups of Gal(E/F), then for each K in ^, the group Gal(E/K) is in &, and for each H in &, the field EH is in ^. Thus, we may define a mapping g: ^ S & by g(K) 5 Gal(E/K) and a mapping f :& S ^ by f (H) 5 EH. It is easy to show that if K and L belong to ^ and K # L, then g(K) $ g(L). Similarly, if G and H belong to & and G # H, then f (G) $ f (H). Thus, f and g are inclusion-reversing mappings between ^ and &. We leave it to the reader to show that for any K in ^, we have (fg)(K) $ K, and for any G in &, we have (gf )(G) $ G. When E is an arbitrary extension of F, these inclusions may be strict. However, when E is a suitably chosen extension of F, the Fundamental Theorem of Galois Theory, Theorem 32.1, says that f and g are inverses

550

Special Topics

of each other, so that the inclusions are equalities. In particular, f and g are inclusion-reversing isomorphisms between the lattices ^ and &. A stronger result than that given in Theorem 32.1 is true, but our theorem illustrates the fundamental principles involved. The student is referred to [1, p. 285] for additional details and proofs. Theorem 32.1 Fundamental Theorem of Galois Theory Let F be a field of characteristic 0 or a finite field. If E is the splitting field over F for some polynomial in F[x], then the mapping from the set of subfields of E containing F to the set of subgroups of Gal(E/F) given by K S Gal(E/K) is a one-to-one correspondence. Furthermore, for any subfield K of E containing F,

1. [E:K] 5 |Gal(E/K)| and [K:F] 5 |Gal(E/F)|/ |Gal(E/K)|. [The index of Gal(E/K) in Gal(E/F) equals the degree of K over F.]

2. If K is the splitting field of some polynomial in F[x],

then Gal(E/K) is a normal subgroup of Gal(E/F) and Gal(K/F) is isomorphic to Gal(E/F)/Gal(E/K). 3. K 5 EGal(E/K). [The fixed field of Gal(E/K) is K.] 4. If H is a subgroup of Gal(E/F), then H 5 Gal(E/EH). [The automorphism group of E fixing EH is H.]

Generally speaking, it is much easier to determine a lattice of subgroups than a lattice of subfields. For example, it is usually quite difficult to determine, directly, how many subfields a given field has, and it is often difficult to decide whether or not two field extensions are the same. The corresponding questions about groups are much more tractable. Hence, the Fundamental Theorem of Galois Theory can be a great labor-saving device. Here is an illustration. [Recall from Chapter 20 that if f (x) [ F[x] and the zeros of f (x) in some extension of F are a1, a2, . . . , an, then F(a1, a2, . . . , an) is the splitting field of f (x) over F.] EXAMPLE 6 Let v 5 cos(2p/7) 1 i sin(2p/7), so that v7 5 1, and consider the field Q(v). How many subfields does it have and what are they? First, observe that Q(v) is the splitting field of x7 2 1 over Q, so that we may apply the Fundamental Theorem of Galois Theory. A simple calculation shows that the automorphism f that sends v to v3 has order 6. Thus, [Q(v):Q] 5 |Gal(Q(v)/Q)| $ 6.

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Also, since x7 2 1 5 (x 2 1)(x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1) and v is a zero of x7 2 1, we see that |Gal(Q(v)/Q)| 5 [Q(v):Q] # 6. Thus, Gal(Q(v)/Q) is a cyclic group of order 6. So, the lattice of subgroups of Gal(Q(v)/Q) is trivial to compute. See Figure 32.6. kφ l 2

3

kφ 2 l

kφ 3 l 3

2 {e}

Figure 32.6 Lattice of subgroups of Gal(Q(v)/Q), where v 5 cos(2p/7) 1 i sin(2p/7)

This means that Q(v) contains exactly two proper extensions of Q: one of degree 3 corresponding to the fixed field of f3 and one of degree 2 corresponding to the fixed field of f2. To find the fixed field of f3, we must find a member of Q(v) that is not in Q and that is fixed by f3. Experimenting with various possibilities leads us to discover that v 1 v21 is fixed by f3 (see Exercise 9), and it follows that Q , Q(v 1 v21) # Q(v)f 3. Since [Q(v)f 3 : Q] 5 3 and [Q(v 1 v21) : Q] divides [Q(v)f 3 : Q], we see that Q(v 1 v21) 5 Q(v)f 3. A similar argument shows that Q(v3 1 v5 1 v6) is the fixed field of f2. Thus, we have found all subfields of Q(v). EXAMPLE 7 Consider the extension E 5 GF(pn) of F 5 GF(p). Let us determine Gal(E/F). By Corollary 2 of Theorem 22.2, E has the form F(b) for some b where b is the zero of an irreducible polynomial p(x) of the form x n 1 an21x n21 1 ? ? ? 1 a1x 1 a0, where an21, an22, . . . , a0 belong to F. Since any field automorphism f of E must take 1 to itself, it follows that f acts as the identity on F. Thus, p(b) 5 0 implies that p(f(b)) 5 0. And because p(x) has at most n zeros, we know that there are at most n possibilities for f(b). On the other hand, by Exercise 45 in Chapter 13, we know that the mapping s (a) 5 a p for all a [ E is an automorphism of E, and it follows from the fact that E* is cyclic (Theorem 22.2) that the group s has order n (see Exercise 11 in Chapter 22). Thus, Gal(GF(p n )/GF(p)) < Zn.

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Special Topics

Solvability of Polynomials by Radicals For Galois, the elegant correspondence between groups and fields given by Theorem 32.1 was only a means to an end. Galois sought to solve a problem that had stymied mathematicians for centuries. Methods for solving linear and quadratic equations were known thousands of years ago (the Quadratic Formula). In the 16th century, Italian mathematicians developed formulas for solving any third- or fourth-degree equation. Their formulas involved only the operations of addition, subtraction, multiplication, division, and extraction of roots (radicals). For example, the equation x3 1 bx 1 c 5 0 has the three solutions A 1 B, 2(A 1 B)/2 1 (A 2 B)"23/2, 2(A 1 B)/2 2 (A 2 B)"23/2, where A5

2c b3 c2 1 1 Ç 2 Å 27 4 3

and

B5

2c b3 c2 2 1 . Ç 2 Å 27 4 3

The formulas for the general cubic x3 1 ax2 1 bx 1 c 5 0 and the general quartic (fourth-degree polynomial) are even more complicated, but nevertheless can be given in terms of radicals of rational expressions of the coefficients. Both Abel and Galois proved that there is no general solution of a fifth-degree equation by radicals. In particular, there is no “quintic formula.” Before discussing Galois’s method, which provided a grouptheoretic criterion for the solution of an equation by radicals and led to the modern-day Galois theory, we need a few definitions. Definition Solvable by Radicals

Let F be a field, and let f (x) [ F[x]. We say that f (x) is solvable by radicals over F if f (x) splits in some extension F(a1, a2, . . . , an) of F and there exist positive integers k1, . . . , kn such that a 1k1 [ F and ai ki [ F(a1, . . . , ai21) for i 5 2, . . . , n.

So, a polynomial in F[x] is solvable by radicals if we can obtain all of its zeros by adjoining nth roots (for various n) to F. In other words,

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each zero of the polynomial can be written as an expression (usually a messy one) involving elements of F combined by the operations of addition, subtraction, multiplication, division, and extraction of roots. 2p "2 "2 EXAMPLE 8 Let v 5 cos( 2p 8 ) 1 i sin( 8 ) 5 2 1 i 2 . Then 8 8 x8 2 3 splits in Q(v, " 3 ), v8 [ Q, and (" 3 )8 [ Q , Q(v). Thus, x8 2 3 is solvable by radicals over Q. Although the zeros of x8 2 3 8 8 8 are most conveniently written in the form " 3 , " 3 v, " 3 v2, . . . , 8 " 3 v7, the notion of solvable by radicals is best illustrated by writing them in the form

6 " 3 , 6 "2 1" 3 , 6 " 3 ("22 1 "212"2), 8

8

8

6 " 3 ("22 2 "212"2). 8

Thus, the problem of solving a polynomial equation for its zeros can be transformed into a problem about field extensions. At the same time, we can use the Fundamental Theorem of Galois Theory to transform a problem about field extensions into a problem about groups. This is exactly how Galois showed that there are fifth-degree polynomials that cannot be solved by radicals, and this is exactly how we will do it. Before giving an example of such a polynomial, we need some additional group theory. Definition Solvable Group

We say that a group G is solvable if G has a series of subgroups {e} 5 H0 , H1 , H2 , ? ? ? , Hk 5 G, where, for each 0 # i , k, Hi is normal in Hi11 and Hi11/Hi is Abelian.

Obviously, Abelian groups are solvable. So are the dihedral groups and any group whose order has the form pn, where p is a prime (see Exercises 22 and 23). The monumental Feit-Thompson Theorem (see Chapter 25) says that every group of odd order is solvable. In a certain sense, solvable groups are almost Abelian. On the other hand, it follows directly from the definitions that any non-Abelian simple group is not solvable. In particular, A5 is not solvable. It follows from Exercise 21 in Chapter 25 that S5 is not solvable. Our goal is to connect the notion of solvability of polynomials by radicals to that of solvable groups. The next theorem is a step in this direction.

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Theorem 32.2 Splitting Field of xn 2 a Let F be a field of characteristic 0 and let a [ F. If E is the splitting field of xn 2 a over F, then the Galois group Gal(E/F) is solvable.

PROOF We first handle the case where F contains a primitive nth root of unity v. Let b be a zero of xn 2 a in E. Then the zeros of xn 2 a are b, vb, v2b, . . . , vn21b, and therefore E 5 F(b). In this case, we claim that Gal(E/F) is Abelian and hence solvable. To see this, observe that any automorphism in Gal(E/F) is completely determined by its action on b. Also, since b is a zero of xn 2 a, we know that any element of Gal(E/F) sends b to another zero of xn 2 a. That is, any element of Gal(E/F) takes b to vib for some i. Let f and s be two elements of Gal(E/F). Then, since v [ F, f and s fix v and f(b) 5 v jb and s(b) 5 v kb for some j and k. Thus, (sf)(b) 5 s (f(b)) 5 s (v jb) 5 s (v j)s (b) 5 v jv kb 5 v j1k b, whereas (fs)(b) 5 f(s(b)) 5 f(v kb) 5 f(v k)f(b) 5 v kv jb 5 v k1jb, so that sf and fs agree on b and fix the elements of F. This shows that sf 5 fs, and therefore Gal(E/F) is Abelian. Now suppose that F does not contain a primitive nth root of unity. Let v be a primitive nth root of unity and let b be a zero of xn 2 a in E. The case where a 5 0 is trivial, so we may assume that b ? 0. Since vb is also a zero of xn 2 a, we know that both v and vb belong to E, and therefore v 5 vb/b is in E as well. Thus, F(v) is contained in E, and F(v) is the splitting field of xn 2 1 over F. Analogously to the case above, for any automorphisms f and s in Gal(F(v)/F) we have f(v) 5 v j for some j and s (v) 5 v k for some k. Then, (sf)(v) 5 s (f(v)) 5 s (v j) 5 (s (v)) j 5 (v k) j 5 (v j ) k 5 (f(v)) k 5 f(v k) 5 f(s (v)) 5 (fs)(v). Since elements of Gal(F(v)/F) are completely determined by their action on v, this shows that Gal(F(v)/F) is Abelian. Because E is the splitting field of xn 2 a over F(v) and F(v) contains a primitive nth root of unity, we know from the case we have already done that Gal(E/F(v)) is Abelian and, by Part 2 of Theorem 32.1, the series {e} # Gal(E/F(v)) # Gal(E/F)

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is a normal series. Finally, since both Gal(E/F(v)) and Gal(E/F)/Gal(E/F(v)) < Gal(F(v)/F) are Abelian, Gal(E/F) is solvable. To reach our main result about polynomials that are solvable by radicals, we need two important facts about solvable groups. Theorem 32.3 Factor Group of a Solvable Group Is Solvable A factor group of a solvable group is solvable.

PROOF Suppose that G has a series of subgroups {e} 5 H0 , H1 , H2 , ? ? ? , Hk 5 G, where, for each 0 # i , k, Hi is normal in Hi11 and Hi11/Hi is Abelian. If N is any normal subgroup of G, then {e} 5 H0N/N , H1N/N , H2N/N , ? ? ? , HkN/N 5 G/N is the requisite series of subgroups that guarantees that G/N is solvable. (See Exercise 25.) Theorem 32.4 N and G/N Solvable Implies G Is Solvable Let N be a normal subgroup of a group G. If both N and G/N are solvable, then G is solvable.

PROOF Let a series of subgroups of N with Abelian factors be N0 , N1 , ? ? ? , Nt 5 N and let a series of subgroups of G/N with Abelian factors be N/N 5 H0 /N , H1/N , ? ? ? , Hs /N 5 G/N. Then the series N0 , N1 , ? ? ? , Nt 5 H0 , H1 , ? ? ? , Hs 5 G has Abelian factors (see Exercise 27). We are now able to make the critical connection between solvability of polynomials by radicals and solvable groups.

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Theorem 32.5 (Galois) Solvable by Radicals Implies Solvable Group Let F be a field of characteristic 0 and let f(x) [ F[x]. Suppose that f(x) splits in F(a1, a2, . . . , at), where a1n1 [ F and aini [ F(a1, . . . , ai21) for i 5 2, . . . , t. Let E be the splitting field for f(x) over F in F(a1, a2, . . . , at ). Then the Galois group Gal(E/F) is solvable.

PROOF We induct on t. For the case t 5 1, we have F # E # F(a1). Let a 5 a1n1 and let L be a splitting field of xn1 2 a over F. Then F # E # L, and both E and L are splitting fields of polynomials over F. By Part 2 of Theorem 32.1, Gal(E/F) < Gal(L/F)/Gal(L/E). It follows from Theorem 32.2 that Gal(L/F) is solvable, and from Theorem 32.3 we know that Gal(L/F)/Gal(L/E) is solvable. Thus, Gal(E/F) is solvable. Now suppose t . 1. Let a 5 a1n1 [ F, let L be a splitting field of xn1 2 a over E, and let K # L be the splitting field of xn1 2 a over F. Then L is a splitting field of (xn1 2 a) f (x) over F, and L is a splitting field of f (x) over K. Since F(a1) # K, we know that f (x) splits in K(a2, . . . , at), so the induction hypothesis implies that Gal(L/K) is solvable. Also, Theorem 32.2 asserts that Gal(K/F) is solvable, which, from Theorem 32.1, tells us that Gal(L/F)/Gal(L/K) is solvable. Hence, Theorem 32.4 implies that Gal(L/F) is solvable. So, by Part 2 of Theorem 32.1 and Theorem 32.3, we know that the factor group Gal(L/F)/Gal(L/E) < Gal(E/F) is solvable. It is worth remarking that the converse of Theorem 32.3 is true also; that is, if E is the splitting field of a polynomial f (x) over a field F of characteristic 0 and Gal(E/F) is solvable, then f (x) is solvable by radicals over F. It is known that every finite group is a Galois group over some field. However, one of the major unsolved problems in algebra, first posed by Emmy Noether, is determining which finite groups can occur as Galois groups over Q. Many people suspect that the answer is “all of them.” It is known that every solvable group is a Galois group over Q. John Thompson has recently proved that certain kinds of simple groups, including the Monster, are Galois groups over Q. The article by Ian Stewart listed among this chapter’s suggested readings provides more information on this topic.

Insolvability of a Quintic We will finish our introduction to Galois theory by explicitly exhibiting a polynomial that has integer coefficients and that is not solvable by radicals over Q.

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Consider g(x) 5 3x5 2 15x 1 5. By Eisenstein’s Criterion (Theorem 17.4), g(x) is irreducible over Q. Since g(x) is continuous and g(22) 5 261 and g(21) 5 17, we know that g(x) has a real zero between 22 and 21. A similar analysis shows that g(x) also has real zeros between 0 and 1 and between 1 and 2. Each of these real zeros has multiplicity 1, as can be verified by long division or by appealing to Theorem 20.6. Furthermore, g(x) has no more than three real zeros, because Rolle’s Theorem from calculus guarantees that between each pair of real zeros of g(x) there must be a zero of g9(x) 5 15x4 2 15. So, for g(x) to have four real zeros, g9(x) would have to have three real zeros, and it does not. Thus, the other two zeros of g(x) are nonreal complex numbers, say, a 1 bi and a 2 bi. (See Exercise 65 in Chapter 15.) Now, let’s denote the five zeros of g(x) by a1, a2, a3, a4, a5. Since any automorphism of K 5 Q(a1, a2, a3, a4, a5) is completely determined by its action on the a’s and must permute the a’s, we know that Gal(K/Q) is isomorphic to a subgroup of S5, the symmetric group on five symbols. Since a1 is a zero of an irreducible polynomial of degree 5 over Q, we know that [Q(a1):Q] 5 5, and therefore 5 divides [K:Q]. Thus, the Fundamental Theorem of Galois Theory tells us that 5 also divides |Gal(K/Q)|. So, by Cauchy’s Theorem (corollary to Theorem 24.3), we may conclude that Gal(K/Q) has an element of order 5. Since the only elements in S5 of order 5 are the 5-cycles, we know that Gal(K/Q) contains a 5-cycle. The mapping from C to C, sending a 1 bi to a 2 bi, is also an element of Gal(K/Q). Since this mapping fixes the three real zeros and interchanges the two complex zeros of g(x), we know that Gal(K/Q) contains a 2-cycle. But, the only subgroup of S5 that contains both a 5-cycle and a 2-cycle is S5. (See Exercise 25 in Chapter 25.) So, Gal(K/Q) is isomorphic to S5. Finally, since S5 is not solvable (see Exercise 21), we have succeeded in exhibiting a fifth-degree polynomial that is not solvable by radicals.

Exercises Seeing much, suffering much, and studying much are the three pillars of learning. BENJAMIN DISRAELI

1. Let E be an extension field of Q. Show that any automorphism of E acts as the identity on Q. (This exercise is referred to in this chapter.) 2. Determine the group of field automorphisms of GF(4). 3. Let E be a field extension of the field F. Show that the automorphism group of E fixing F is indeed a group. (This exercise is referred to in this chapter.)

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4. Given that the automorphism group of Q("2, "5, "7) is isomorphic to Z2 % Z2 % Z2, determine the number of subfields of Q("2, "5, "7) that have degree 4 over Q. 5. Let E be a field extension of a field F and let H be a subgroup of Gal(E/F). Show that the fixed field of H is indeed a field. (This exercise is referred to in this chapter.) 6. Let E be the splitting field of x4 1 1 over Q. Find Gal(E/Q). Find all subfields of E. Find the automorphisms of E that have fixed fields Q("2), Q("2 2), and Q(i). Is there an automorphism of E whose fixed field is Q? 7. Let f (x) [ F[x] and let the zeros of f (x) be a1, a2, . . . , an. If K 5 F(a1, a2, . . . , an), show that Gal(K/F) is isomorphic to a group of permutations of the ai’s. [When K is the splitting field of f (x) over F, the group Gal(K/F) is called the Galois group of f (x).] 8. Show that the Galois group of a polynomial of degree n has order dividing n!. 9. Referring to Example 6, show that the automorphism f has order 6. Show that v 1 v21 is fixed by f3 and v3 1 v5 1 v6 is fixed by f2. (This exercise is referred to in this chapter.) 10. Let E 5 Q("2, "5). What is the order of the group Gal(E/Q)? What is the order of Gal(Q("10)/Q)? 11. Suppose that F is a field of characteristic 0 and E is the splitting field for some polynomial over F. If Gal(E/F) is isomorphic to A4, show that there is no subfield K of E such that [K:F] 5 2. 12. Determine the Galois group of x3 2 1 over Q and x3 2 2 over Q. (See Exercise 7 for the definition.) 13. Suppose that K is the splitting field of some polynomial over a field F of characteristic 0. If [K:F] 5 p2q, where p and q are distinct primes, show that K has subfields L1, L2, and L3 such that [K:L1] 5 p, [K:L2] 5 p2, and [K:L3] 5 q. 14. Suppose that E is the splitting field of some polynomial over a field F of characteristic 0. If Gal(E/F) is isomorphic to D5, draw the subfield lattice for the fields between E and F. 15. Suppose that F , K , E are fields and E is the splitting field of some polynomial in F[x]. Show, by means of an example, that K need not be the splitting field of some polynomial in F[x]. 16. Suppose that E is the splitting field of some polynomial over a field F of characteristic 0. If [E:F] is finite, show that there is only a finite number of fields between E and F. 17. Suppose that E is the splitting field of some polynomial over a field F of characteristic 0. If Gal(E/F) is an Abelian group of order 10, draw the subfield lattice for the fields between E and F.

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18. Let v be a nonreal complex number such that v5 5 1. If f is the automorphism of Q(v) that carries v to v4, find the fixed field of f. 19. Determine the isomorphism class of the group Gal(GF(64)/GF(2)). 20. Determine the isomorphism class of the group Gal(GF(729)/GF(9)). Exercises 21, 22, and 23 are referred to in this chapter. 21. 22. 23. 24. 25.

Show that S5 is not solvable. Show that the dihedral groups are solvable. Show that a group of order pn, where p is prime, is solvable. Show that Sn is solvable when n # 4. Complete the proof of Theorem 32.3 by showing that the given series of groups satisfies the definition for solvability. 26. Show that a subgroup of a solvable group is solvable. 27. Let N be a normal subgroup of G and let K/N be a normal subgroup of G/N. Prove that K is a normal subgroup of G. (This exercise is referred to in this chapter.) 28. Show that any automorphism of GF( pn) acts as the identity on GF( p).

Reference 1. G. Ehrlich, Fundamental Concepts of Abstract Algebra, Boston: PWS-Kent, 1991.

Suggested Readings Tony Rothman, “The Short Life of Évariste Galois,” Scientific American, April (1982): 136–149. This article gives an elementary discussion of Galois’s proof that the general fifth-degree equation cannot be solved by radicals. The article also goes into detail about Galois’s controversial life and death. In this regard, Rothman refutes several accounts given by other Galois biographers. Ian Stewart, “The Duelist and the Monster,” Nature 317 (1985): 12–13. This nontechnical article discusses recent work of John Thompson pertaining to the question of “which groups can occur as Galois groups.”

Suggested Website To find more information about the history of quadratic, cubic, and quartic equations, visit:

http://www-groups.dcs.st-and.ac.uk/~history/

Philip Hall He [Hall] was preeminent as a group theorist and made many fundamental discoveries; the conspicuous growth of interest in group theory in the 20th century owes much to him. J. E. ROSEBLADE

PHILIP HALL was born on April 11, 1904, in London. Abandoned by his father shortly after birth, Hall was raised by his mother, a dressmaker. He demonstrated academic prowess early by winning a scholarship to Christ’s Hospital, where he had several outstanding mathematics teachers. At Christ’s Hospital, Hall won a medal for the best English essay, the gold medal in mathematics, and a scholarship to King’s College, Cambridge. Although abstract algebra was a field neglected at King’s College, Hall studied Burnside’s book Theory of Groups and some of Burnside’s later papers. After graduating in 1925, he stayed on at King’s College for further study and was elected to a fellowship in 1927. That same year, Hall discovered a major “Sylow-like” theorem about solvable groups: If a solvable group has order mn, where m and n are relatively prime, then every subgroup whose order divides m is contained in a group of order m and all subgroups of order m are conjugate. Over the next three decades, Hall developed a general theory of

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finite solvable groups that had a profound influence on John Thompson’s spectacular achievements of the 1960s. In the 1930s, Hall also developed a general theory of groups of prime-power order that has become a foundation of modern finite group theory. In addition to his fundamental contributions to finite groups, Hall wrote many seminal papers on infinite groups. Among the concepts that have Hall’s name attached to them are Hall subgroups, Hall algebras, Hall-Littlewood polynomials, Hall divisors, the marriage theorem from graph theory, and the Hall commutator collecting process. Beyond his own discoveries, Hall had an enormous influence on algebra through his research students. No fewer than one dozen have become eminent mathematicians in their own right. Hall died on December 30, 1982. To find more information about Hall, visit: http://www-groups.dcs.st-and .ac.uk/~history/

33 Cyclotomic Extensions

“. . . to regard old problems from a new angle requires creative imagination and marks real advances in science.” ALBERT EINSTEIN

Motivation For the culminating chapter of this book, it is fitting to choose a topic that ties together results about groups, rings, fields, geometric constructions, and the history of mathematics. The so-called cyclotomic extensions is such a topic. We begin with the history. The ancient Greeks knew how to construct regular polygons of 3, 4, 5, 6, 8, 10, 12, 15, and 16 sides with a straightedge and compass. And, given a construction of a regular n-gon, it is easy to construct a regular 2n-gon. The Greeks attempted to fill in the gaps (7, 9, 11, 13, 14, 17, . . .) but failed. More than 2200 years passed before anyone was able to advance our knowledge of this problem beyond that of the Greeks. Incredibly, Gauss, at age 19, showed that a regular 17-gon is constructible, and shortly thereafter he completely solved the problem of exactly which n-gons are constructible. It was this discovery of the constructibility of the 17-sided regular polygon that induced Gauss to dedicate his life to the study of mathematics. Gauss was so proud of this accomplishment that he requested that a regular 17-sided polygon be engraved on his tombstone. Gauss was led to his discovery of the constructible polygons through his investigation of the factorization of polynomials of the form x n 2 1 over Q. In this chapter, we examine the factors of x n 2 1 and show how Galois theory can be used to determine which regular n-gons are constructible with a straightedge and compass. The irreducible factors of x n 2 1 are important in number theory and combinatorics.

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Cyclotomic Polynomials Recall from Example 2 in Chapter 16 that the complex zeros of xn 2 1 2p 2 3 n21. Thus, the splitting are 1, v 5 cos(2p n ) 1 i sin( n ), v , v , . . . , v n field of x 2 1 over Q is Q(v). This field is called the nth cyclotomic extension of Q, and the irreducible factors of x n 2 1 over Q are called the cyclotomic polynomials. 2p Since v 5 cos(2p n ) 1 i sin( n ) generates a cyclic group of order n under multiplication, we know from Corollary 3 of Theorem 4.2 that the generators of v are the elements of the form vk, where 1 # k # n and gcd(n, k) 5 1. These generators are called the primitive nth roots of unity. Recalling that we use f(n) to denote the number of positive integers less than or equal to n and relatively prime to n, we see that for each positive integer n there are precisely f(n) primitive nth roots of unity. The polynomials whose zeros are the f(n) primitive nth roots of unity have a special name. Definition

For any positive integer n, let v1, v2, . . . , vf(n) denote the primitive nth roots of unity. The nth cyclotomic polynomial over Q is the polynomial Fn(x) 5 (x 2 v1)(x 2 v2) ? ? ? (x 2 vf(n)).

In particular, note that Fn(x) is monic and has degree f(n). In Theorem 33.2 we will prove that Fn(x) has integer coefficients, and in Theorem 33.3 we will prove that Fn(x) is irreducible over Z. EXAMPLE 1 F1(x) 5 x 2 1, since 1 is the only zero of x 2 1. F2(x) 5 x 1 1, since the zeros of x 2 2 1 are 1 and 21, and 21 is the only primitive 2p root. F3(x) 5 (x 2 v)(x 2 v2), where v 5 cos(2p 3 ) 1 i sin( 3 ) 5 (21 1 i"3)/2, and direct calculations show that F3(x) 5 x2 1 x 1 1. Since the zeros of x4 2 1 are 61 and 6i and only i and 2i are primitive, F4(x) 5 (x 2 i)(x 1 i) 5 x2 1 1. In practice, one does not use the definition of Fn(x) to compute it. Instead, one uses the formulas given in the exercises and makes recursive use of the following result. Theorem 33.1 For every positive integer n, xn 2 1 5 Pd|nFd (x), where the product runs over all positive divisors d of n.

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Before proving this theorem, let us be sure that the statement is clear. For n 5 6, for instance, the theorem asserts that x6 2 1 5 F1(x)F2(x)F3(x)F6(x), since 1, 2, 3, and 6 are the positive divisors of 6. PROOF Since both polynomials in the statement are monic, it suffices to show that they have the same zeros and that all zeros have multi2p plicity 1. Let v 5 cos(2p n ) 1 i sin( n ). Then v is a cyclic group of order n, and v contains all the nth roots of unity. From Theorem 4.3 we know that for each j, |v j| divides n so that (x 2 v j) appears as a factor in F|v j |(x). On the other hand, if x 2 a is a linear factor of Fd (x) for some divisor d of n, then a d 5 1, and therefore a n 5 1. Thus, x 2 a is a factor of x n 2 1. Finally, since no zero of x n 2 1 can be a zero of Fd (x) for two different d’s, the result is proved. Before we illustrate how Theorem 33.1 can be used to calculate Fn(x) recursively, we state an important consequence of the theorem. Theorem 33.2 For every positive integer n, Fn(x) has integer coefficients.

PROOF The case n 5 1 is trivial. By induction, we may assume that g(x) 5 Pd|nFd (x) has integer coefficients. From Theorem 33.1 we d,n

know that x n 2 1 5 Fn(x)g(x), and, because g(x) is monic, we may carry out the division in Z[x] (see Exercise 45 in Chapter 16). Thus, Fn(x) [ Z[x]. Now let us do some calculations. If p is a prime, we have from Theorem 33.1 that x p 2 1 5 F1(x)Fp(x) 5 (x 2 1)Fp(x), so that Fp(x) 5 (x p 2 1)/(x 2 1) 5 x p21 1 x p22 1 ? ? ? 1 x 1 1. From Theorem 33.1 we have x6 2 1 5 F1(x)F2(x)F3(x)F6(x), so that F6(x) 5 (x6 2 1)/((x 2 1)(x 1 1)(x2 1 x 1 1)). So, by long division, F6(x) 5 x2 2 x 1 1. Similarly, F10(x) 5 (x10 2 1)/ ((x 2 1)(x 1 1)(x4 1 x3 1 x2 1 x 1 1)) 5 x4 2 x3 1 x2 2 x 1 1. The exercises provide shortcuts that often make long division unnecessary. The values of Fn(x) for all n up to 15 are shown in Table 33.1. The software for Computer Exercise 1 provides the values for Fn(x) for all values of n up to 1000. Judging from Table 33.1, one might be led to conjecture that 1 and 21 are the only nonzero coefficients of the

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Table 33.1 The Cyclotomic Polynomials Fn(x) up to n 5 15 n

Fn(x)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

x21 x11 x2 1 x 1 1 x2 1 1 x4 1 x3 1 x2 1 x 1 1 x2 2 x 1 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1 x4 1 1 x6 1 x3 1 1 x4 2 x3 1 x2 2 x 1 1 x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1 x4 2 x2 1 1 x12 1 x11 1 x10 1 x9 1 x8 1 x7 1 x6 1 x5 1 x4 1 x3 1 x2 1 x 1 1 x6 2 x5 1 x4 2 x3 1 x2 2 x 1 1 x8 2 x7 1 x5 2 x4 1 x3 2 x 1 1

cyclotomic polynomials. However, it has been shown that every integer is a coefficient of some cyclotomic polynomial. The next theorem reveals why the cyclotomic polynomials are important. Theorem 33.3 (Gauss) The cyclotomic polynomials Fn(x) are irreducible over Z.

PROOF Let f (x) [ Z[x] be a monic irreducible factor of Fn(x). Because Fn(x) is monic and has no multiple zeros, it suffices to show that every zero of Fn(x) is a zero of f(x). Since Fn(x) divides xn 2 1 in Z[x], we may write xn 2 1 5 f (x)g(x), where g(x) [ Z[x]. Let v be a primitive nth root of unity that is a zero of f (x). Then f (x) is the minimal polynomial for v over Q. Let p be any prime that does not divide n. Then, by Corollary 3 of Theorem 4.2, v p is also a primitive nth root of unity, and therefore 0 5 (v p)n 2 1 5 f (v p)g(v p), and so f (v p) 5 0 or g(v p) 5 0. Suppose f (v p) 2 0. Then g(v p) 5 0, and so v is a zero of g(x p). Thus, from Theorem 21.3, f (x) divides g(x p) in Q[x]. Since f (x) is monic, f (x) actually divides g(x p) in Z[x] (see Exercise 45 in Chapter 16). Say g(x p) 5 f (x)h(x), where h(x) [ Z[x]. Now let g(x), f (x), and h(x) denote the polynomials in Z p[x] obtained from g(x), f (x), and h(x), respectively, by reducing each coefficient modulo p. Since this reduction process is a ring homomorphism from Z[x] to Z p[x] (see Exercise 9 in Chapter 16), we

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have g(x p) 5 f (x)h(x) in Z p[x]. From Exercise 31 in Chapter 16 and Corollary 5 of Theorem 7.1, we then have ( g(x)) p 5 g(x p) 5 f (x)h(x), and since Z p[x] is a unique factorization domain, it follows that g(x) and f (x) have an irreducible factor in Zp[x] in common; call it m(x). Thus, we may write f (x) 5 k1(x)m(x) and g(x) 5 k2(x)m(x), where k1(x), k2(x) [ Zp[x]. Then, viewing xn 2 1 as a member of Zp[x], we have xn 2 1 5 f (x)g(x) 5 k1(x)k2(x)(m(x))2. In particular, x n 2 1 has a multiple zero in some extension of Z p. But because p does not divide n, the derivative nx n21 of x n 21 is not 0, and so nx n21 and x n 2 1 do not have a common factor of positive degree in Zp[x]. Since this contradicts Theorem 20.5, we must have f (v p) 5 0. We reformulate what we have thus far proved as follows: If b is any primitive nth root of unity that is a zero of f (x) and p is any prime that does not divide n, then b p is a zero of f (x). Now let k be any integer between 1 and n that is relatively prime to n. Then we can write k 5 p1 p2 ? ? ? pt, where each pi is a prime that does not divide n (repetitions are permitted). It follows then that each of v, v p1, (v p1) p2, . . . , (v p1 p2???pt–1) pt 5 vk is a zero of f (x). Since every zero of Fn(x) has the form v k, where k is between 1 and n and is relatively prime to n, we have proved that every zero of Fn(x) is a zero of f (x). This completes the proof. Of course, Theorems 33.3 and 33.1 give us the factorization of x n 2 1 as a product of irreducible polynomials over Q. But Theorem 33.1 is also useful for finding the irreducible factorization of x n 2 1 over Zp. The next example provides an illustration. Irreducible factors of x n 2 1 over Zp are used to construct error-correcting codes. EXAMPLE 2 We determine the irreducible factorization of x6 2 1 over Z2 and Z3. From Table 33.1, we have x6 2 1 5 (x 2 1)(x 1 1)(x 2 1 x 1 1)(x 2 2 x 1 1). Taking all the coefficients on both sides mod 2, we obtain the same expression, but we must check that these factors are irreducible over Z2. Since x 2 1 x 1 1 has no zeros in Z2, it is irreducible over Z2 (see Theorem 17.1). Finally, since 21 5 1 in Z2, we have the irreducible factorization x 6 2 1 5 (x 1 1)2 (x 2 1 x 1 1)2. Over Z3, we again start with the factorization x6 2 1 5 (x 2 1)(x 1 1)(x 2 1 x 1 1)(x 2 2 x 1 1) over Z and view the coefficients mod 3. Then 1 is a zero of x 2 1 x 1 1 in Z 3, and by long division we obtain x 2 1 x 1 1 5 (x 2 1)(x 1 2) 5 (x 1 2)2. Similarly, x 2 2 x 1 1 5 (x 2 2)(x 1 1) 5 (x 1 1)2. So, the irreducible factorization of x 6 2 1 over Z 3 is (x 1 1)3 (x 1 2)3. We next determine the Galois group of the cyclotomic extensions of Q.

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Theorem 33.4 Let v be a primitive nth root of unity. Then Gal(Q(v)/Q) < U(n).

PROOF Since 1, v, v2, . . . , v n21 are all the nth roots of unity, Q(v) is the splitting field of x n 2 1 over Q. For each k in U(n), v k is a primitive nth root of unity, and by the lemma preceding Theorem 20.4, there is a field automorphism of Q(v), which we denote by fk, that carries v to v k and acts as the identity on Q. Moreover, these are all the automorphisms of Q(v), since any automorphism must map a primitive nth root of unity to a primitive nth root of unity. Next, observe that for every r, s [ U(n), (fr fs )(v) 5 fr (v s ) 5 (fr (v)) s 5 (v r ) s 5 v rs 5 f rs (v). This shows that the mapping from U(n) onto Gal(Q(v)/Q) given by k S f k is a group homomorphism. Clearly, the mapping is an isomorphism, since v r 2 v s when r, s [ U(n) and r 2 s. The next example uses Theorem 33.4 and the results of Chapter 8 to demonstrate how to determine the Galois group of cyclotomic extensions. 2p 2p EXAMPLE 3 Let a 5 cos(2p 9 ) 1 i sin( 9 ) and let b 5 cos( 15 ) 1 2p i sin( 15 ). Then

Gal(Q(a)/Q) < U(9) < Z6 and Gal(Q(b)/Q) < U(15) < U(5) % U(3) < Z4 % Z2.

The Constructible Regular n-gons As an application of the theory of cyclotomic extensions and Galois theory, we determine exactly which regular n-gons are constructible with a straightedge and compass. But first we prove a technical lemma. Lemma 2p Let n be a positive integer and let v 5 cos A 2p n B 1 i sin A n B . Then QAcos A 2p n BB # Q(v). 2p 2p 2p PROOF Observe that from (cos(2p n ) 1 i sin( n ))(cos( n ) 2 i sin( n )) 5 2p 2p 2p 2p cos2( n ) 1 sin2( n ) 5 1, we have cos( n ) 2 i sin( n ) 5 1/v. Moreover, 2p 2p (v 1 1/v)/2 5 (2 cos(2p n ))/2 5 cos( n ). Thus, cos( n ) [ Q(v).

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Theorem 33.5 (Gauss, 1796) It is possible to construct the regular n-gon with a straightedge and compass if and only if n has the form 2kp1 p2 ? ? ? pt, where k $ 0 and the pi’s are distinct primes of the form 2m 1 1.

PROOF If it is possible to construct a regular n-gon, then we can construct the angle 2p/n and therefore the number cos(2p n ). By the results of Chapter 23, we know that cos(2p ) is constructible only if [Q(cos(2p n n )) : Q] is a power of 2. To determine when this is so, we will use Galois theory. 2p Let v 5 cos(2p n ) 1 i sin( n ). Then |Gal(Q(v)/Q)| 5 [Q(v) : Q] 5 f(n). By the lemma on the preceding page, Q(cos(2p n )) # Q(v), and by Theorem 32.1 we know that 2p [Q(cos(2p n )) : Q] 5 |Gal(Q(v)/Q)|/|Gal(Q(v)/Q(cos( n )))| 2p 5 f(n)/|Gal(Q(v)/Q(cos( n )))|.

Recall that the elements s of Gal(Q(v)/Q) have the property that 2p 2pk s(v) 5 vk for 1 # k # n. That is, s(cos(2p n ) 1 i sin( n )) 5 cos( n ) 1 2pk 2p i sin( n ). If such a s belongs to Gal(Q(v)/Q(cos( n ))), then we must 2p have cos(2pk n ) 5 cos( n ). Clearly, this holds only when k 5 1 and k 5 2p n 2 1. So, |Gal(Q(v)/Q(cos(2p n )))| 5 2, and therefore [Q(cos( n )) : Q] 5 f(n)/2. Thus, if an n-gon is constructible, then f(n)/2 must be a power of 2. Of course, this implies that f(n) is a power of 2. Write n 5 2kp1n 1p2 n 2 ? ? ? pt n t, where k $ 0, the pi’s are distinct odd primes, and the ni’s are positive. Then f(n)5|U(n)| 5 |U(2k)||U( p1n1)| |U( p2n2)|? ? ? |U( pt n t)| 5 2k21p1n 121( p1 2 1) p2n221 ( p2 2 1) ? ? ? pt n t21( pt 2 1) must be a power of 2. Clearly, this implies that each ni 5 1 and each pi 2 1 is a power of 2. This completes the proof that the condition in the statement is necessary. To prove that the condition given in Theorem 33.5 is also sufficient, suppose that n has the form 2k p1p2 ? ? ? pt , where the pi’s are distinct 2p odd primes of the form 2m 1 1, and let v 5 cos(2p n ) 1 i sin( n ). By Theorem 33.3, Q(v) is a splitting field of an irreducible polynomial over Q, and therefore, by the Fundamental Theorem of Galois Theory, f(n) 5 [Q(v) : Q] 5 |Gal(Q(v)/Q)|. Since f(n) is a power of 2 and Gal(Q(v)/Q) is an Abelian group, it follows by induction (see Exercise 15) that there is a series of subgroups H0 , H1 , ? ? ? , Ht 5 Gal(Q(v)/Q) where H0 is the identity, H1 is the subgroup of Gal(Q(v)/Q) of order 2 that fixes cos(2p n ), and |Hi11 : Hi| 5 2 for i 5 0, 1, 2, . . . , t 2 1. By the

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Fundamental Theorem of Galois Theory, we then have a series of subfields of the real numbers Q 5 EH , EH t

t–1

, ? ? ? , EH 5 Q(cos(2p n )), 1

where [EH : EH ] 5 2. So, for each i, we may choose bi [ EH such i21 i i21 that EH 5 EH (bi). Then bi is a zero of a polynomial of the form i21 i x 2 1 bi x 1 ci [ EH [x], and it follows that EH 5 EH ("b2i 2 4ci). i i21 i Thus, it follows from Exercise 3 in Chapter 23 that every element of Q(cos(2p n )) is constructible. It is interesting to note that Gauss did not use Galois theory in his proof. In fact, Gauss gave his proof 15 years before Galois was born.

Exercises Difficulties should act as a tonic. They should spur us to greater exertion. B. C. FORBES

1. Determine the minimal polynomial for cos( p3 ) 1 i sin( p3 ) over Q. 2. Factor x12 2 1 as a product of irreducible polynomials over Z. 3. Factor x 8 2 1 as a product of irreducible polynomials over Z2, Z3, and Z5. 4. For any n . 1, prove that the sum of all the nth roots of unity is 0. 5. For any n . 1, prove that the product of the nth roots of unity is (21)n11. 6. Let v be a primitive 12th root of unity over Q. Find the minimal polynomial for v4 over Q. 7. Let F be a finite extension of Q. Prove that there are only a finite number of roots of unity in F. 8. For any n . 1, prove that the irreducible factorization over Z of x n21 1 xn22 1 ? ? ? 1 x 1 1 is PFd (x), where the product runs over all positive divisors d of n greater than 1. 9. If 2n 1 1 is prime for some n $ 1, prove that n is a power of 2. (Primes of the form 2n 1 1 are called Fermat primes.) 10. Prove that Fn(0) 5 1 for all n . 1. 11. Prove that if a field contains the nth roots of unity for n odd, then it also contains the 2nth roots of unity. 12. Let m and n be relatively prime positive integers. Prove that the splitting field of x mn 2 1 over Q is the same as the splitting field of (x m 2 1)(x n 2 1) over Q.

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13. Prove that F2 n(x) 5 Fn(2x) for all odd integers n . 1. 14. Prove that if p is a prime and k is a positive integer, then Fp k (x) 5 Fp(x p k 21). Use this to find F8(x) and F27(x). 15. Prove the assertion made in the proof of Theorem 33.5 that there exists a series of subgroups H0 , H1 , ? ? ? , Ht with |Hi11 : Hi| 5 2 for i 5 0, 1, 2, . . . , t 2 1. (This exercise is referred to in this chapter.) 16. Prove that x 9 2 1 and x 7 2 1 have isomorphic Galois groups over Q. (See Exercise 7 in Chapter 32 for the definition.) 17. Let p be a prime that does not divide n. Prove that Fpn(x) 5 Fn(x p)/ Fn(x). 18. Prove that the Galois groups of x 10 2 1 and x 8 2 1 over Q are not isomorphic. 19. Let E be the splitting field of x 5 2 1 over Q. Show that there is a unique field K with the property that Q , K , E. 20. Let E be the splitting field of x 6 2 1 over Q. Show that there is no field K with the property that Q , K , E. 2p 21. Let v 5 cos(2p 15 ) 2 i sin( 15 ). Find the three elements of Gal(Q(v)/Q) of order 2.

Computer Exercises Mathematics is not a deductive science—that’s a cliché. When you try to prove a theorem, you don’t just list the hypotheses, and then start to reason. What you do is trial and error, experimentation, guesswork. PAUL HALMOS, I Want to Be a Mathematician.

Software for the first computer exercise in this chapter is available at the website: http://www.d.umn.edu/~jgallian 1. This program returns the nth cyclotomic polynomial. Enter several choices for n of the form pq and p2q, where p and q are distinct primes. Make a conjecture about the nonzero coefficients of Fn(x). 2. Use computer software such as Mathematica, Maple, or GAP to find the irreducible factorization over Z of all polynomials of the form x n 2 1, where n is between 2 and 100. On the basis of this information, make a conjecture about the nature of coefficients of the irreducible factors of x n 2 1 for all n. Then test your conjecture for n 5 105.

Carl Friedrich Gauss He [Gauss] lives everywhere in mathematics. E. T. BELL, Men of Mathematics

CARL FRIEDRICH GAUSS, considered by many to be the greatest mathematician who has ever lived, was born in Brunswick, Germany, on April 30, 1777. While still a teenager, he made many fundamental discoveries. Among these were the method of “least squares” for handling statistical data, and a proof that a 17-sided regular polygon can be constructed with a straightedge and compass (this result

This stamp was issued by East Germany in 1977. It commemorates Gauss’s construction of a regular 17-sided polygon with a straightedge and compass.

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was the first of its kind since discoveries by the Greeks 2000 years earlier). In his Ph.D. dissertation in 1799, he proved the Fundamental Theorem of Algebra. Throughout his life, Gauss largely ignored the work of his contemporaries and, in fact, made enemies of many of them. Young mathematicians who sought encouragement from him were usually rebuffed. Despite this fact, Gauss had many outstanding students, including Eisenstein, Riemann, Kummer, Dirichlet, and Dedekind. Gauss died in Göttingen at the age of 77 on February 23, 1855. At Brunswick, there is a statue of him. Appropriately, the base is in the shape of a 17-point star. In 1989, Germany issued a bank note (see page 112) depicting Gauss and the Gaussian distribution. To find more information about Gauss, visit: http://www-groups.dcs .st-and.ac.uk/~history/

Manjul Bhargava We are watching him [Bhargava] very closely. He is going to be a superstar. He's amazingly mature mathematically. He is changing the subject in a fundamental way. PETER SARNAK

MANJUL BHARGAVA was born in Canada on August 8, 1974, and grew up in Long Island, New York. After graduating from Harvard in 1996, Bhargava went to Princeton to pursue his Ph.D. under the direction of Andrew Wiles (see page 340). Bhargava investigated a “composition law” first formulated by Gauss in 1801 for combining two quadratic equations (equations in a form such as x2 1 3xy 16y2 5 0) in a way that was very different from normal addition and revealed a lot of information about number systems. Bhargava tackled an aspect of the problem in which no progress had been made in more than 200 years. He not only broke new ground in that area but also discovered 13 more composition laws and developed a coherent mathematical framework to explain them. He then applied his theory of composition to solve a number of fundamental problems concerning the distribution of extension fields of the rational numbers and of other, related algebraic objects. What made Bhargava’s work especially remarkable is that he was able to explain all his revolutionary ideas using only elementary mathematics. In commenting on Bhargava’s results, Wiles said, “He did it in a way that

Gauss himself could have understood and appreciated.” Despite his youth, Bhargava already has won many awards, including a Clay Research Fellowship, the Clay Research Award, the Blumenthal Award for the Advancement of Research in Pure Mathematics, the SASTRA Ramanujan Prize, and the 2008 Cole Prize in number theory (see page 430). In 2002 he was named one of Popular Science magazine’s “Brilliant 10,” in celebration of scientists who are shaking up their fields. In 2003, Bhargava accepted a full professorship with tenure at Princeton at the age of 28. In addition to doing mathematics, Bhargava is an accomplished tabla player who has studied with the world’s most distinguished tabla masters. He performs extensively in the New York and Boston areas. To hear him play the tabla, visit http://www.npr.org/templates/story/ story.php?storyId=4111253 To find more information about Bhargava, visit www.Wikipedia.org and www.d.umn.edu/~jgallian/ manjulMH4.pdf

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Supplementary Exercises for Chapters 24–33

Text not available due to copyright restrictions

True/false questions for Chapters 24–33 are available on the Web at http://www.d.umn.edu/~jgallian/TF 1. Let G 5 x, y | x 5 (xy)3, y 5 (xy)4. To what familiar group is G isomorphic? 2. Let G 5 z | z6 5 1 and H 5 x, y | x2 5 y3 5 1, xy 5 yx. Show that G and H are isomorphic. 3. Show that a group of order 315 5 32 ? 5 ? 7 has a subgroup of order 45. 4. Let G be a group of order p 2q 2, where p and q are primes and p . q. If |G| 2 36, prove that G has a normal Sylow p-subgroup. 5. Let H denote a Sylow 7-subgroup of a group G and K a Sylow 5-subgroup of G. Assume that |H| 5 49, |K| 5 5, and K is a subgroup of N(H). Show that H is a subgroup of N(K). 6. Prove that no finite group of order greater than 6 has exactly three conjugacy classes. 7. Suppose that K is a normal Sylow p-subgroup of H and that H is a normal subgroup of G. Prove that K is a normal subgroup of G. (Compare this with Exercise 53 in Chapter 9.) 8. Show that the polynomial x 5 2 6x 1 3 over Q is not solvable by radicals. 9. Let H and K be subgroups of G. Prove that HK is a subgroup of G if H # N(K). 10. Suppose that H is a subgroup of a finite group G and that H contains N(P), where P is some Sylow p-subgroup of G. Prove that N(H) 5 H. 11. Prove that a simple group G of order 168 cannot contain an element of order 21. 12. Prove that the only group of order 561 is Z561.

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13. Prove that the center of a non-Abelian group of order 105 has order 5. 14. Let n be an odd integer that is at least 3. Prove that every Sylow subgroup of Dn is cyclic. 15. Let G be the digraph obtained from Cay({(1, 0), (0, 1)}: Z3 % Z5) by deleting the vertex (0, 0). [Also, delete each arc to or from (0, 0).] Prove that G has a Hamiltonian circuit. 16. Prove that the digraph obtained from Cay({(1, 0), (0, 1)}: Z4 % Z7) by deleting the vertex (0, 0) has a Hamiltonian circuit. 17. Let G be a finite group generated by a and b. Let s1, s2, . . . , sn be the arcs of a Hamiltonian circuit in the digraph Cay({a, b}: G). We say that the vertex s1s2 ? ? ? si travels by a if si11 5 a. Show that if a vertex x travels by a, then every vertex in the coset xab21 travels by a. 18. Recall that the dot product u ? v of two vectors u 5 (u1, u2, . . . , un) and v 5 (v1, v2, . . . , vn) from F n is u1v1 1 u2v2 1 ? ? ? 1 unvn (where the addition and multiplication are those of F). Let C be an (n, k) linear code. Show that C > 5 {v [ Fn | v ? u 5 0 for all u [ C} 19.

20. 21. 22. 23.

is an (n, n 2 k) linear code. This code is called the dual of C. Find the dual of each of the following binary codes: a. {00, 11}, b. {000, 011, 101, 110}, c. {0000, 1111}, d. {0000, 1100, 0011, 1111}. Let C be a binary linear code such that C # C >. Show that wt(v) is even for all v in C. Let C be an (n, k) binary linear code. If v is a binary n-tuple, but v o C >, show that v ? u 5 0 for exactly half of the elements u in C. Suppose that C is an (n, k) binary linear code and the vector 11 ? ? ? 1 [ C >. Show that wt(v) is even for every v in C. Suppose that C is an (n, k) binary linear code and C 5 C >. (Such a code is called self-dual.) Prove that n is even. Prove that 11 ? ? ? 1 is a code word.

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24. If G is a finite solvable group, show that there exist subgroups of G {e} 5 H0 , H1 , H2 , ? ? ? , Hn 5 G such that Hi11/Hi has prime order.

The End. Title of song by JOHN LENNON AND PAUL MCCARTNEY, Abbey Road, side 2, October 1969

Selected Answers Don’t wait for answers Just take your chances Don’t ask me why BILLY JOEL, “Don’t Ask Me Why,” Glass Houses

Many of the proofs given below are merely sketches. In these cases, the student should supply the complete proof. Chapter 0 To make headway, improve your head. B. C. FORBES

1. {1, 2, 3, 4}; {1, 3, 5, 7}; {1, 5, 7, 11}; {1, 3, 7, 9, 11, 13, 17, 19}; {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24} 3. 12, 2, 2, 10, 1, 0, 4, 5. 5. 1942, June 18; 1953, December 13. 7. By using 0 as an exponent if necessary, we may write a 5 p1m1 . . . pkmk and b 5 p1n1 . . . pknk, where the p’s are distinct primes and the m’s and n’s are nonnegative. Then lcm (a, b) 5 p1s1 . . . pksk, where si 5 max(mi, ni), and gcd(a, b) 5 p1t1 . . . pktk, where ti 5 min(mi, ni). Then lcm(a, b) ? gcd(a, b) 5 p1m11n1 . . . pkmk1nk 5 ab. 9. Write a 5 nq1 1 r1 and b 5 nq2 1 r2, where 0 # r1, r2 , n. We may assume that r1 $ r2. Then a 2 b 5 n(q1 2 q2) 1 (r1 2 r2), where r1 2 r2 $ 0. If a mod n 5 b mod n, then r1 5 r2 and n divides a 2 b. If n divides a 2 b, then by the uniqueness of the remainder, we have r1 2 r2 5 0. 11. Use Exercise 9. 13. Use the “GCD Is a Linear Combination” theorem (Theorem 0.2). 15. Let p be a prime greater than 3. By the Division Algorithm, we can write p in the form 6n 1 r, where r satisfies 0 # r , 6. Now observe that 6n, 6n 1 2, 6n 1 3, and 6n 1 4 are not prime. 17. Since st divides a 2 b, both s and t divide a 2 b. The converse is true when gcd(s, t) 5 1. 19. Use Euclid’s Lemma and the Fundamental Theorem of Arithmetic. 21. Use proof by contradiction. 23. Let S be a set with n 1 1 elements and pick some a in S. By induction, S has 2n subsets that do not contain a. But there is a one-to-one correspondence between the subsets of S that do not contain a and those that do. So, there are 2 ? 2n 5 2n11 subsets in all. 25. Consider n 5 200! 1 2. 27. Say p1p2 . . . pr 5 q1q2 . . . qs, where the p’s and q’s are primes. By the Generalized Euclid’s Lemma, p1 divides some qi, say q1 (we may relabel the q’s if necessary). Then p1 5 q1 and p2 . . . pr 5 q2 . . . qs. Repeating this argument at each step, we obtain p2 5 q2, . . . , pr 5 qr and r 5 s. 29. Suppose that S is a set that contains a and whenever n $ a belongs to S, then n 1 1 [ S. We must prove that S contains all integers greater than or equal to a. Let T be the set of all integers greater than a that are not in S and suppose that T is not empty. Let b be the smallest integer in T (if T has no negative integers, b exists because of the Well Ordering Principle; if T has negative integers, it can have only a finite number of them so that there is a smallest one). Then b 2 1 [ S, and therefore b 5 (b 2 1) 1 1 [ S. A1

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31. The statement is true for any divisor of 84 2 4 5 4092. 33. 6 P.M. 35. Observe that the number with the decimal representation a9a8 . . . a1a0 is a9 ? 109 1 a8 ? 108 1 . . . 1 a1 ? 10 1 a0. Then use Exercise 11 and the fact that ai10i mod 9 5 ai mod 9 to deduce that the check digit is (a9 1 a8 1 . . . 1 a1 1 a0) mod 9. 37. For the case in which the check digit is not involved, see the answer to Exercise 35. If a transposition involving the check digit c 5 (a1 1 a2 1 . . . 1 a10) mod 9 goes undetected, then a10 5 (a1 1 a2 1 . . . 1 a 9 1 c) mod 9. Substitution yields 2(a1 1 a2 1 . . . 1 a9 ) mod 9 5 0. Therefore, modulo 9, we have 10(a1 1 a2 1 . . . 1 a 9) 5 a1 1 a2 1 . . . 1 a9 5 0. It follows that c 5 a10. In this case the transposition does not yield an error. 39. Say the number is a8a7 . . . a1a0 5 a8 ? 108 1 a7 ? 107 1 . . . 1 a1 ? 10 1 a0. Then the error is undetected if and only if (ai10i 2 ai910i) mod 7 5 0. Multiplying both sides by 5i and noting that 50 mod 75 1, we obtain (ai 2 ai9) mod 7 5 0. 41. 4 45. Cases where (2a 2 b 2 c) mod 11 5 0 are undetected. 47. The check digit would be the same. 49. 4302311568 51. 2. Since b is one-to-one, b(a (a1)) 5 b(a(a2)) implies that a(a1) 5 a(a2) and since a is one-toone, a1 5 a2. 3. Let c [ C. There is a b in B such that b(b) 5 c and an a in A such that a(a) 5 b. Thus, (ba)(a) 5 b(a(a)) 5 b(b) 5 c. 4. Since a is one-to-one and onto we may define a21(x) 5 y if and only if a( y) 5 x. Then a21(a(a)) 5 a and a(a21(b)) 5 b. 53. No. (1, 0) [ R and (0, 21) [ R, but (1, 21) o R. 55. a belongs to the same subset as a. If a and b belong to the subset A, then b and a also belong to A. If a and b belong to the subset A and b and c belong to the subset B, then A 5 B, since the distinct subsets of P are disjoint. So, a and c belong to A. 57. Apply g 21 to both sides of ag 5 bg.

Chapter 1 Think of what you’re saying, you can get it wrong and still think that it’s all right. JOHN LENNON AND PAUL MCCARTNEY,

“We Can Work It Out,” single

1. Three rotations: 08, 1208, 2408, and three reflections across lines from vertices to midpoints of opposite sides. 3. no 5. Dn has n rotations of the form k(3608/n), where k 5 0, . . . , n 2 1. In addition, Dn has n reflections. When n is odd, the axes of reflection are the lines from the vertices to the midpoints of the opposite sides. When n is even, half of the axes of reflection are obtained by joining opposite vertices; the other half, by joining midpoints of opposite sides. 7. A rotation followed by a rotation either fixes every point (and so is the identity) or fixes only the center of rotation. However, a reflection fixes a line. 9. Observe that 1 ? 1 5 1; 1(21) 5 21; (21)1 5 21; (21)(21) 5 1. These relationships also hold when 1 is replaced by “rotation” and 21 is replaced by “reflection.” 11. HD 5 DV but H 2 V. 13. R0, R180, H, V 15. See answer for Exercise 13. 17. In each case, the group is D6. 19. cyclic 21. Their only symmetry is the identity. 23. It would wobble violently.

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Chapter 2 The noblest pleasure is the joy of understanding. LEONARDO DA VINCI

1. Does not contain the identity; closure fails. 3. Under modulo 4, 2 does not have an inverse. Under modulo 5, each element has an inverse. 9 9 d 10 8 a. 2a 1 3b; b. 22a 1 2(2b 1 c); c. 23(a 1 2b) 1 2c 5 0 e Use the fact that det (AB) 5 (det A)(det B). 29 (ab)n need not equal anbn in a non-Abelian group. Use the Socks-Shoes Property. For the case n . 0, use induction. For n , 0, note that e 5 (a21ba)n(a21ba)2n 5 (a21ba)n (a21b2na) and solve for (a21ba)n. {1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45} Suppose x appears in a row labeled with a twice. Say x 5 ab and x 5 ac; then cancellation yields b 5 c. But we use distinct elements to label the columns. Use Exercise 23. a21cb21; aca21 If x3 5 e and x 2 e, then (x21)3 5 e and x 2 x21. So, nonidentity solutions come in pairs. If x2 2 e, then x21 2 x and (x21)2 2 e. So solutions to x2 2 e come in pairs. Observe that since RF is a reflection, we have (RF)(RF ) 5 R0. So, RFR 5 F21 5 F. Observe that aa21b 5 ba21a. Since a2 5 b2 5 (ab)2 5 e, we have aabb 5 abab. Now cancel on the left and right. If n is not prime, the set is not closed under multiplication modulo n. If n is prime, the set is closed and for every r in the set there are integers s and t such that 1 5 rs 1 nt 5 rs modulo n. 1>2 1>2 Closure follows from the definition of multiplication. The identity is c d . The inverse of 1>2 1>2 a a 1> (4a) 1> (4a) c d is c d. a a 1> (4a) 1> (4a)

5. c 7. 9. 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39.

Chapter 3 The brain is as strong as its weakest think. ELEANOR DOAN

1. |Z12| 5 12; |U(10)| 5 4; |U(12)| 5 4; |U(20)| 5 8; |D4| 5 8 In Z12, |0| 5 1; |1| 5 |5| 5 |7| 5 |11| 5 12; |2| 5 |10| 5 6; |3| 5 |9| 5 4; |4| 5 |8| 5 3; |6| 5 2. In U(10), |1| 5 1; |3| 5 |7| 5 4; |9| 5 2. In U(12), |1| 5 1; |5| 5 2; |7| 5 2; |11| 5 2. In U(20), |1| 5 1; |3| 5 |7| 5 |13| 5 |17| 5 4; |9| 5 |11| 5 |19| 5 2. In D4, |R0| 5 1; |R90| 5 |R270| 5 4; |R180| 5 |H| 5 |V| 5 |D| 5 |D9| 5 2. In each case, notice that the order of the element divides the order of the group. 3. In Q, |0| 5 1 and all other elements have infinite order. In Q*, |1| 5 1, |21| 5 2, and all other elements have infinite order. 5. Each is the inverse of the other. 7. Suppose that m , n and am 5 an. Then e 5 ana 2m 5 an2m. This contradicts the assumption that a has infinite order.

A4

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9. If a has infinite order, then e, a, a2, . . . are all distinct and belong to G, so G is infinite. If |a| 5 n, then e, a, a2, . . . , an 2 1 are distinct and belong to G. 11. By brute force, show that k 4 5 1 for all k. 13. For any integer n $ 3, Dn contains elements a and b of order 2 with |ab| 5 n. In general, there is no relationship among |a|, |b|, and |ab|. 15. 2,3,6. 17. U4(20) 5 {1, 9, 13, 17}; U5(20) 5 {1, 11}; U5(30) 5 {1, 11}; U10(30) 5 {1, 11}. Uk(n) is closed because (ab) mod k 5 (a mod k)(b mod k) 5 1 ? 1 5 1. H is not closed. 19. If x [ Z(G), then x [ C(a) for all a, so x [ > C(a). If x [ > C(a), then xa 5 ax for all a in G, a[G a[G so x [ Z(G). 21. The case that k 5 0 is trivial. Let x [ C(a). If k is positive, then by induction on k, xak11 5 xaak1 axak 5 aakx 5 ak11x. The case where k is negative now follows from Exercise 20. 23. a. C(5) 5 G; C(7) 5 {1, 3, 5, 7} b. Z(G) 5 {1, 5} c. |2| 5 2; |3| 5 4. They divide the order of the group. 25. Mimic the proof of Theorem 3.5. 27. No. In D4, C(R180) 5 D4. 29. For the first part, see Example 4. For the second part, use D4. 31. Since the only elements of finite order in R* are 1 and 21, the only finite subgroups are {1} and {1, 21}. 33. 2 35. First observe that (a d)n/d 5 an 5 e, so 0 ad 0 is at most n/d. Moreover, there is no positive integer t , n>d such that (ad ) t 5 adt 5 e, for otherwise 0a 0 ? n. 1 1 n 1 n 37. Note that c d 5 c d. 0 1 0 1 39. For any positive integer n, a rotation of 3608/n has order n. A rotation of "2° has infinite order. 41. R0, R90, R180, D, D9, H, V. (Note that R90 5 R270). The subgroups {R0, R180, D, D9} and {R0, R180, H, V} are not cyclic. 43. Nonidentity elements of odd order come in pairs. So, there must be some element a of even order, say |a| 5 2m. Then |am| 5 2. 45. Let |g| 5 m and write m 5 nq 1 r where 0 # r , n. Then gr 5 gm2nq 5 gm(gn)2q 5 (gn)2q belongs to H. So, r 5 0. 47. 1 [ H. Let a, b [ H. Then (ab21)2 5 a2(b2)21, which is the product of two rationals. 2 can be replaced by any positive integer. 49. |3| 5 4 a b a9 b9 51. Let c d and c d belong to H. It suffices to show that a 2 a9 1 b 2 b9 1 c 2 c9 1 d 2 c d c9 d9 d9 5 0. This follows from a 1 b 1 c 1 d 5 0 5 a9 1 b9 1 c9 1 d9. If 0 is replaced by 1, H is not a subgroup. 53. If 2a and 2b [ K, then 2a(2b)21 5 2a2b [ K, since a 2 b [ H. 1 2 0 21 0 55. c d 5 c 2 1 d is not in H. 0 2 0 2 57. If a 1 bi and c 1 di [ H, then (a 1 bi)(c 1 di)21 5 (ac 1 bd) 1 (bc 2 ad)i and (ac 1 bd)2 1 (bc 2 ad)2 5 1, so that H is a subgroup. H is the unit circle in the complex plane.

a1b a d 0 ab 1 b2 2 a2; a, b [ R f a b a b 2 b. e c d 0 a 2 b2; a, b [ R f b a a 0 c. e c d |a 2 0; a [ R f 0 a

59. a. e c

61. Use Theorem 0.2.

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A5

Chapter 4 There will be an answer, let it be. JOHN LENNON AND PAUL MCCARTNEY, “Let It Be,” single

1. For Z6, generators are 1 and 5; for Z8, generators are 1, 3, 5, and 7; for Z20, generators are 1, 3, 7, 9, 11, 13, 17, and 19. 3. 20 5 520, 10, 06; 10 5 510, 20, 06 a20  5 5a20, a10, a06; a10  5 5a10, a20, a06 5.  3  5 {3, 9, 7, 1};  7  5 {7, 9, 3, 1} 7. U(8) or D3. 9. Six subgroups; generators are the divisors of 20. Six subgroups; generators are ak, where k is a divisor of 20. 11. By definition, a21 [ a. So, a 21  # a. By definition, a 5 (a 21 ) 21 [ a 21 . So, a # a 21 . 13. 21 > 10 5 18 5 6. In the general case am  > an  5 ak , where k 5 1cm (m, n) mod 24. 15. |g| divides 12 is equivalent to g12 5 e. So, if a12 5 e and b12 5 e, then (ab21)12 5 a12(b12)21 5 ee21 5 e. The general result is given in Exercise 29 of Chapter 3. 17. is odd or infinite 19. 1, 7, 11, 17, 19, 29 21. a. |a| divides 12. b. |a| divides m. c. By Theorem 4.3, |a| 5 1, 2, 3, 4, 6, 8, 12, or 24. If |a| 5 2, then a8 5 (a2)4 5 e4 5 e. A similar argument eliminates all other possibilities except 24. 23. Yes, by Theorem 4.3. The subgroups of Z are of the form n 5 {0, 6n, 62n, 63n, . . .}, n 5 0, 1, 2, 3, . . . . The subgroups of a are of the form an  for n 5 0, 1, 2, 3, . . . . 25. For the first part, use Theorem 4.4; Dn has n elements of order 2 when n is odd and n 1 1 elements of order 2 when n is even. 27. See Example 14 of Chapter 2. 29. 1000000, 3000000, 5000000, 7000000. By Theorem 4.3,  1000000  is the unique subgroup of order 8, and only those on the list are generators; a1000000, a3000000, a5000000, a7000000. By Theorem 4.3, a1000000  is the unique subgroup of order 8, and only those on the list are generators. 31. Let G 5 {a1, a2, . . . , ak}. Now let |ai| 5 ni. Consider n 5 n1n2 . . . nk. 33. Mimic Exercise 32. 35. Mimic Exercise 34. 37. Suppose a and b are relatively prime positive integers and a/b 5 Q1. Then there is some positive integer n such that (a/b)n 5 2. Clearly, n 2 0, 1, or 21. If n . 1, an 5 2bn, so that 2 divides a. But then 2 divides b as well. A similar contradiction occurs if n , 21. 39. For 6, use Z 25. For n, use Z 2n21. 41. Let t 5 lcm(m, n) and |ab| 5 s. Then (ab)t 5 atbt 5 e, and therefore s divides t. Also, e 5 (ab)s 5 asbs, so that as 5 b2s, and therefore as and b2s belong to a > b 5 {e}. Thus, m divides s and n divides s, and, therefore, t divides s. This proves that s 5 t. For the second part, try D3. 43. An infinite cyclic group does not have an element of prime order. A finite cyclic group can have only one subgroup for each divisor of its order. A subgroup of order p has exactly p 2 1 elements of order p. Another element of order p would give another subgroup of order p. 45. 1 ? 4, 3 ? 4, 7 ? 4, 9 ? 4; x 4,(x 4 ) 3, (x 4 ) 7, (x 4 ) 9. 47. 1 of order 1; 33 of order 2; 2 of order 3; 10 of order 11; 20 of order 33 49. 1, 2, 10, 20 51. Say a and b are distinct elements of order 2. If a and b commute, then ab is a third element of order 2. If a and b do not commute, then aba is a third element of order 2. 53. Use Exercise 18 of Chapter 3 and Theorem 4.3. 55. 1 and 2 57. In a cyclic group there are at most n solutions to the equation xn 5 e. 59. 12 or 60; 48

A6

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61. Observe that a280 5 e 5 a440. Thus 0a 0 is common divisor of 280 and 440, and therefore 0a 0 divides gcd(280, 440) 5 40. 63. Say b is a generator of the group. Since p and pn 2 1 are relatively prime, we know by Corollary 3 of Theorem 4.2 that bp also generates the group. Finally, observe that (b p)k 5 (b k) p. 65. Use the fact that a cyclic group of even order has a unique element of order 2. 67. G is a group because it is closed. It is not cyclic because every nonzero element has order 3. 69. Since m and n are relatively prime, it suffices to show both m and n divide k. By Corollary 2 of Theorem 4.1, it is enough to show that ak 5 e. Note that ak [ a > b, and since a > b is a subgroup of both a and b, we know that 0 a > b 0 must divide both 0 a 0 and 0 b 0 . Thus, 0 a > b 0 5 1. 71. Observe that among the integers from 1 to pn the pn21 integers p, 2p, 3p, . . . , pn21p are exactly the ones that are not relatively prime to p.

Supplementary Exercises for Chapters 1–4 Four short words sum up what has lifted most successful individuals above the crowd: a little bit more. They did all that was expected of them and a little bit more. A. LOU VICKERY

1. a. Let xh1x21 and xh2x21 belong to xHx21. Then (xh1x21)(xh2x21)21 5 xh1h221x21 [ xHx21 also. b. Let h 5 H. Then xhx21 5 xHx21. c. (xh1x21)(xh2x21) 5 xh1h2x21 5 xh2h1x21 5 (xh2x21)(xh1x21) 3. Suppose cl(a) > cl(b) 2 f. Say xax21 5 yby21. Then (y21x)a(y21x)21 5 b. Thus, for any ubu21 in cl(b), we have ubu21 5 (uy21x)a(uy21x)21 [ cl(a). This shows that cl(b) # cl(a). By symmetry, cl(a) # cl(b). Because a 5 eae21 [ cl(a), the union of the conjugacy classes is G. 5. Observe that (xax21)k 5 xakx21. Thus, (xax21)k 5 e if and only if ak 5 e. 7. Try D4. 9. By Exercise 5, for every x in G, |xax21| 5 |a|, so that xax21 5 a or xa 5 ax. 11. 1 of order 1, 15 of order 2, 8 of order 15, 4 of order 5, 2 of order 3. 13. Let |G| 5 5. Let a 2 e belong to G. If |a| 5 5, we are done. If |a| 5 3, then {e, a, a2} is a subgroup of G. Let b be either of the remaining two elements of G. Then the set {e, a, a2, b, ab, a2b} consists of six different elements, a contradiction. Thus, |a| 2 3. Similarly, |a| 2 4. We may now assume that every nonidentity element of G has order 2. Pick a 2 e and b 2 e in G with a 2 b. Then {e, a, b, ab} is a subgroup of G. Let c be the remaining element of G. Then {e, a, b, ab, c, ac, bc, abc} is a set of eight distinct elements of G, a contradiction. It now follows that if a [ G and a 2 e, then |a| 5 5. 15. an(bn)21 5 (ab21)n, so Gn is a subgroup. For the non-Abelian group, try D3. 17. Suppose G 5 H < K. Pick h [ H with h o K. Pick k [ K, but k o H. Then, hk [ G, but hk o H and hk o K. U(8) is the union of the three subgroups. 19. If |a| 5 pk and |b| 5 pr with k # r, say, then |ab21| divides pr. 21. Note that ba2 5 ab and a3 5 b2 5 e imply ba 5 a2b. Thus, every member of the group can be written in the form a i b j. Therefore, the group is {e, a, a2, b, ab, a2b}. D3 satisfies these conditions. 23. xy 5 yx if and only if xyx21y21 5 e. But, (xy)x21y21 5 x21(xy)y21 5 ee 5 e. 25. Let x [ N(gHg21). Then x(gHg21)x21 5 gHg21. Thus g21xgHg21x21g 5 g21xgH(g21xg)21 5 H. This means that g21xg [ N(H). So x [ gN(H)g21. Reverse the argument to show gN(H)g21 # N(gHg21). 27. Look at D11. 29. Solution from Mathematics Magazine.† “Yes. Let a be an arbitrary element of S. The set {an | n 5 1, 2, 3, . . .} is finite, and therefore a m 5 a n for some m, n with m . n 9 1. By cancellation we

†Mathematics

Magazine 63 (April 1990): 136.

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31. 33. 35. 37.

39.

41. 43. 45.

47.

A7

have ar(a) 5 a, where r(a) 5 m 2 n 1 1 . 1. If x is any element of S, then aar(a)21x 5 ar(a)x 5 ax, and this implies that ar(a)21x 5 x. Similarly, we see that xar(a)21 5 x, and the element e 5 ar(a)21 is an identity. The identity element is unique, for if e9 is another identity, then e 5 ee9 5 e9. If r(a) . 2 then ar(a)22 is an inverse of a, and if r(a) 5 2 then a2 5 a 5 e is its own inverse. Thus S is a group.” 11 is rational so H 2 f. Say am and bn are rational. Then (ab21)mn 5 (am)n/(bn)m is rational. Use det (AB) 5 (det A)(det B) to prove H is a subgroup. H is not a subgroup when det A is an integer, since det A21 need not be an integer. Choose x 2 e and y o x. Then G 5 x < y. But then xy [ y, so that x # y and therefore G 5 y. To prove that |G| 5 pq or p3, use Theorem 4.3. If T and U are not closed, then there are elements x and y in T and w and z in U such that xy is not in T and wz is not in U. It follows that xy [ U and wz [ T. Then xywz 5 (xy)wz [ U and xywz 5 xy(wz) [ T, a contradiction. Let G be the group of all polynomials with integer coefficients under addition. Let Hk be the subgroup of polynomials of degree at most k together with the zero polynomial (the zero polynomial does not have a degree). Take g 5 a. Let S 5 {s1, s2, s3, . . . , sk} and let g be any element in G. Then the set {gs121, gs221, gs321, . . . , gsk21} and S have at least one element in common. Say gsi21 5 sj. Then g 5 sj si. Let K 5 {x [ G | |x| divides d}. The sub test shows that K is a subgroup. Let x [ H. By Theorem 4.3, |x| divides d. So, H # K. Let y [ K, |y| 5 t, and d 5 tq. By Theorem 4.3, H has a subgroup of order t and G has only one subgroup of order t. So, y # H. To check associativity, note (a * b) * c 5 ((a 1 b) 2 1) * c 5 a 1 b 2 1 1 c 2 1 5 a 1 b 1 c 2 2 and a * (b * c) 5 a * (b 1 c 2 1) 5 a 1 (b 1 c 2 1) 2 1 5 a 1 b 1 c 2 2. To determine the identity e, we observe that a * e 5 a if and only if a 1 e 2 1 5 a. Thus 1 is the identity (it is obvious that the operation is commutative). If a21 exists, we have must a * a21 5 a 1 a21 2 1 5 1, and therefore a21 is 2a 1 2. To find a generator, observe that for any positive integer k, ak 5 ka 2 (k 2 1). So, for positive k and a 5 2, we have 2k 5 k 1 1. One can also check that 2k 5 k 1 1 when k 5 0 or negative. Thus 2 generates all integers.

Chapter 5 Mistakes are often the best teachers. JAMES A. FROUDE

1. 3. 5. 7. 9. 11. 13.

a. 2 b. 3 c. 5 a. 3 b. 12 c. 6 d. 6 e. 12 f. 2 12 For S6, the possible orders are 1, 2, 3, 4, 5, 6; for A6, 1, 2, 3, 4, 5; for A7, 1, 2, 3, 4, 5, 6, 7. a. even b. odd c. even d. odd e. even even; odd An even number of 2-cycles followed by an even number of 2-cycles gives an even number of 2-cycles in all. So the finite subgroup test is verified. 15. Suppose that a can be written as a product of m 2-cycles and b can be written as a product of n 2-cycles. Then ab can be written as a product of m 1 n 2-cycles. Now observe that m 1 n is even if and only if m and n are both even or both odd. 1 2 3 4 5 6 17. a. a21 5 c d 2 1 3 5 4 6 1 2 3 4 5 6 b. ba 5 c d 1 6 2 3 4 5 1 2 3 4 5 6 c. ab 5 c d 6 2 1 5 3 4

A8

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19. Suppose H contains at least one odd permutation, say, s. Imitate the proof of Theorem 5.7 with s in place of (12). 21. The identity is even; the set is not closed. 23. a. C(a3) 5 {a1, a2, a3, a4}; b. C(a12) 5 {a1, a7, a12} 25. 180; 75 27. b 5 (2457136) 29. (124586739), (142568793), (214856379). 31. Let a, b [ stab(a). Then ab(a) 5 a(b(a)) 5 a(a) 5 a. Also, a(a) 5 a implies a21(a(a)) 5 a21(a) or a 5 a21(a). 33. m is a multiple of 6 but not a multiple of 30. 35. 6!/5 5 144 37. 3, 7, 9 39. Let a 5 (123) and b 5 (145). 41. (123)(12) 2 (12)(123) in Sn (n $ 3). 43. Cycle decomposition shows that any nonidentity element of A5 is a 5-cycle, a 3-cycle, or a product of a pair of disjoint 2-cycles. Then, observe that there are (5 ? 4 ? 3 ? 2 ? 1)/5 5 24 group elements of the form (abcde), (5 ? 4 ? 3)/3 5 20 group elements of the form (abc), and (5 ? 4 ? 3 ? 2)/(2 ? 2 ? 2) 5 15 group elements of the form (ab)(cd). 45. One possibility is {(1), (12)(34), (56)(78), (12)(34)(56)(78)}. 47. Hint: (13)(12) 5 (123) and (12)(34) 5 (324)(132). 49. Verifying that a * s(b) 2 b * s(a) is done by examining all cases. To prove the general case, observe that si(a) * s i11(b) 2 s i(b) * s i11(a) can be written in the form si(a) * s(si(b)) 2 si(b) * s(si(a)), which is the case already done. If a transposition were not detected, then s(a1) * . . . * si(ai) * si11(ai11) * . . . * sn(an) 5 s(a1) * . . . * si(ai11) * si11(ai) * . . . * sn(an), which implies si(ai) * s i11(ai11) 5 si(ai11) * s i11(ai). 51. Observe that (a1a2 . . . an) 5 (1a1)(1an)(1an21) . . . (1a1). 53. If a has odd order k and a is an odd permutation, then e 5 ak would be odd. 55. By case-by-case analysis, H is a subgroup for n 5 1, 2, 3 and 4. For n $ 5, observe that (12)(34) and (12)(35) belong to H but their product does not. 57. The product of an element of Z(A4) of order 2 and an element of A4 of order 3 would have order 6. The product of an element of Z(A4) of order 3 and an element of A4 of order 2 would have order 6. 59. Labeling the four tires 1, 2, 3, and 4 in clockwise order starting with 1 being the tire in the upper left-hand corner, we may represent the four patterns as a 5 (1324) top left-hand pattern b 5 (1423) top right-hand pattern g 5 (14)(23) bottom right-hand pattern d 5 (13)(24) bottom left-hand pattern Notice that a21 5 b and that d 5 a2g. Thus, we need only find the smallest subgroup of S4 containing a and g. To this end, observe that the set {e, a, a2, a3, g, ag, a2g, a3g} is closed under multiplication on the left and right by both a and g. This implies that the set is closed under multiplication and is therefore a group. Since ag 2 ga, the subgroup is non-Abelian.

Chapter 6 Think and you won’t sink. B. C. FORBES, Epigrams

1. 3. 5. 7.

Try n → 2n. f(xy) 5 "xy 5 "x "y 5 f(x)f(y) Try 1 → 1, 3 → 5, 5 → 7, 7 → 11. D12 has elements of order 12 and S4 does not.

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A9

9. Since Te(x) 5 ex 5 x for all x, Te is the identity. For the second part, observe that Tg 8 (Tg)21 5 Te 5 Tgg21 5 Tg 8 Tg21 and cancel. 11. For any x in the group, we have (fgfh)(x) 5 fg(fh(x)) 5 fg(hxh21) 5 ghxh21g21 5 (gh)x(gh)21 5 fgh(x). 13. fR and fR disagree on H; fR and fH disagree on R90; fR and fD disagree on R90. The remain90 0 90 90 ing cases are similar. 15. Let a [ Aut(G). We show that a21 is operation-preserving: a21(xy) 5 a21(x)a21(y) if and only if a(a21(xy)) 5 a(a21(x)a21(y)), that is, if and only if xy 5 a(a21(x))a(a21(y)) 5 xy. So a21 is operation-preserving. That Inn(G) is a group follows from the equation fgfh 5 fgh. 17. That a is one-to-one follows from the fact that r21 exists modulo n. The operation-preserving condition is Exercise 11 in Chapter 0. 19. Use Part 2 of Theorem 6.2. 21. The inverse of a one-to-one function is one-to-one. To see that f21 is operation-preserving, let a and b belong to G. Then f21(ab) 5 f21(a) f21(b) if and only if ab 5 f(f21(a))f(f21(b)) 5 ab [we obtained the first equality by applying f to both sides of f21(ab) 5 f21(a)f21(b)]. Finally, let g [ G. Then f21(f(g)) 5 g, so that f21 is onto. 23. Tg(x) 5 Tg(y) if and only if gx 5 gy or x 5 y. This shows that Tg is a one-to-one function. Let y [ G. Then Tg(g21y) 5 y, so that Tg is onto. 25. Apply the appropriate definitions. 27. Show that Q is not cyclic. a 2b 29. Try a 1 bi → c d. b a 31. Yes, by Cayley’s Theorem. 33. Observe that fg(y) 5 gyg21 and fzg(y) 5 zgy(zg)21 5 zgyg21z21 5 gyg21 since z [ Z(G). So, fg 5 fzg. 35. fg 5 fh implies gxg21 5 hxh21 for all x. This implies h21gx(h21g)21 5 x, and therefore h21g [ Z(G). 37. Say |a| 5 n. Then f a n(x) 5 anxa2n 5 x, so that f an is the identity. For the example, take a 5 R90 in D4. 39. Observe that D 5 R90V and H 5 R90D. 41. (R0R90R180R270)(H D9V D). 43. Consider the mapping f(x) 5 x2 and note that 2 is not in the image. 45. Use the fact that if a . 0, then a 5 "a"a. For the second part, use the first part together with the fact that the inverse of an automorphism is an automorphism. 47. Say f is an isomorphism from Q to R1 and f takes 1 to a. It follows that the integer r maps to ar and the rational r/s maps to ar/s. But ar/s 2 ap for any r/s.

Chapter 7 Use missteps as stepping stones to deeper understanding and greater achievement. SUSAN TAYLOR

1. 3. 5. 7. 9. 11.

H 5 {a1, a2, a3, a4}, a5H 5 {a5, a8, a6, a7}, a9H 5 {a9, a11, a12, a10} H, 1 1 H, 2 1 H a. yes b. yes c. no 8/2 5 4, so there are four cosets. Let H 5 {1, 11}. The cosets are H, 7H, 13H, 19H. Since |a4| 5 15, there are two cosets: a4 and aa4. Suppose that h [ H and h , 0. Then hR1 # hH 5 H. But hR1 is the set of all negative real numbers. Thus, H 5 R*. 13. 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 15. Use Lagrange’s Theorem (Theorem 7.1) and Corollary 3. 17. By Exercise 16, we have 56 mod 7 5 1 . So, using mod 7, we have 515 5 56 ? 56 ? 52 ? 5 5 1 ? 1 ? 4 ? 5 5 6; 713 mod 11 5 2.

A10

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19. Use Corollary 4 of Lagrange’s Theorem (Theorem 7.1) together with Theorem 0.2. 21. By closure (234)(12) 5 (1342) belongs to H so that |H| is divisible by 3 and 4 and divides 24. But if |H| 5 12 then the even permutations in H would be a subgroup of A4 of order 6, which does not exist (see Example 5). 23. Since G has odd order, no element can have order 2. Thus, for each x 2 e, we know that x 2 x21. So, we can write the product of all the elements in the form ea1a121a2 a221 . . . an a n21 5 e. 25. Let H be the subgroup of order p and K be the subgroup of order q. Then H < K has p 1 q 2 1 , pq elements. Let a be any element in G that is not in H < K. By Lagrange’s Theorem, |a| 5 p, q, or pq. But |a| 2 p, for if so, then a 5 H. Similarly, |a| 2 q. 27. 1, 3, 11, 33. If |x| 5 33, then |x11| 5 3. Elements of order 11 occur in multiples of 10. 29. No. Observe that by Lagrange’s Theorem, the elements of a group of order 55 must have orders 1, 5, 11, or 55; then use Theorem 4.4. 31. Observe that |G:H| 5 |G|/|H|, |G:K| 5 |G|/|K|, and |K:H| 5 |K|/|H|. 33. Certainly, a [ orbG(a). Now suppose that c [ orbG(a) > orbG(b). Then c 5 a(a) and c 5 b(b) for some a and b, and therefore (b21a)(a) 5 b. So, if x [ orbG(b), then x 5 g(b) 5 (gb21a)(a) for some g. This proves that orbG(b) # orbG(a). By symmetry, orbG(a) # orbG(b). 35. a. stabG(1) 5 {(1), (24)(56)}; orbG(1) 5 {1, 2, 3, 4} b. stabG(3) 5 {(1), (24)(56)}; orbG(3) 5 {3, 4, 1, 2} c. stabG(5) 5 {(1), (12)(34), (13)(24), (14)(23)}; orbG(5) 5 {5, 6} 37. Suppose that |Z(G)| 5 pn21 and let a be an element of G not in Z(G). Then C(a) contains both a and Z(G). By Lagrange’s Theorem, we must have C(a) 5 G. But then a [ Z(G). 39. 2520 41. Consider the mapping from G to G defined by f(x) 5 x2 and let |G| 5 2k 1 1. Use the observation that x 5 xe 5 xx2k11 5 x2k12 5 (x2)k11 to prove that f is one-to-one and Exercise 10 of Chapter 5 to show that f is onto. 43. Suppose that B [ G and det(B) 5 2. Then det(A21B) 5 1, so that A21B [ H and therefore B [ AH. Conversely, for any Ah [ AH we have det(Ah) 5 det(A)det(h) 5 2 ? 1 5 2. 45. It is the set of all permutations that carry face 2 to face 1. 47. aH 5 bH if and only if det (a) 5 6det (b). 49. 50

Chapter 8 There is always a right and a wrong way, and the wrong way always seems the more reasonable. GEORGE MOORE, The Bending of The Bough

1. Closure and associativity in the product follow from the closure and associativity in each component. The identity in the product is the n-tuple with the identity in each component. The inverse of (g1, g2, . . . , gn) is (g121, g221, . . . , gn21). 3. Use g → (g, eH) and h → (eG, h). 5. To show that Z % Z is not cyclic, note that (a, b 1 1) o (a, b). 7. Use (g1, g2) → (g2, g1). In general, G1 % G2 . . . % Gn is isomorphic to the external direct product of any rearrangement of G1,G2, . . . ,Gn. 9. Yes, by Theorem 8.2. 11. Observe by Theorem 4.4 that as long as d divides n, the number of elements of order d in a cyclic group depends only on d. So, in both Z8000000 and Z4 there are f(4) 5 2 elements of order 4 and f(2) 5 1 element of order 2. Similarly for Zm % Zn. 13. Try a 1 bi → (a, b). 15. Use Exercise 3 and Theorem 4.3. 17. m>r % n>s.

A11

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A11

19. Since (g, h) # g % h, a necessary and sufficient condition for equality is that lcm( |g|,|h|) 5 | (g,h)| 5 |g % h| 5 |g| |h| . This is equivalent to gcd( |g|,|h|) 5 1 . 21. | (a,b,c)| 5 lcm5 |a|, |b|, |c|6 5 3 , unless a 5 b 5 c 5 e. In general, the order of every nonidentity element of Zp % Zp % . . . % Zp, where p is prime, is p. a b 23. Map c d to (a, b, c, d). Let Rk denote c d R % R % ? ? ? % R (k factors). Then the group of m 3 n matrices under addition is isomorphic to Rmn. 25. (g, g)(h, h)21 5 (gh21, gh21) When G 5 R, G % G is the plane and H is the line y 5 x. 27. (3, 0), (3, 1), (3, 2), (0, 1) 29. 60 31. {0, 400} % {0, 50, 100, 150} 33. Compare the number of elements of order 2 in each group. 35. The mapping f(3m6n ) 5 (m, n) is an isomorphism. The mapping f(3m9n ) 5 (m, n) is not well-defined since f(3290 ) 5 f(30, 91 ) . 37. D24 has elements of order 24, whereas D3 % D4 does not. 39. 12 41. Aut(U(25)) < Aut(Z20) < U(20) < U(4) % U(5) < Z2 % Z4. 43. 2k 2 1; 2t 2 1, where t is the number of the n1, n2, . . . , nk that are even. 45. No. Z10 % Z12 % Z6 has 7 elements of order 2 whereas Z15 % Z4 % Z12 has only 3. 47. Using the fact that an isomorphism from Z12 is determined by the image of 1 and the fact that a generator must map to a generator, we determine that there are 4 isomorphisms. 49. Since a [ Zm and b [ Zn, we know that |a| divides m and |b| divides n. So, |(a, b)| 5 lcm(|a|, |b|) divides lcm(m, n). 51. Z, Z3, Z4, Z6 53. Observe that every nonidentity element of Zp % Zp has order p and each subgroup of order p contains p 2 1 of them. So, there are exactly (p2 2 1)/(p 2 1) 5 p 1 1 subgroups of order p. 55. Look at Z % Z2. 57. U(165) < U(11) % U(15) < U(5) % U(33) < U(3) % U(55) < U(3) % U(5) % U(11) 59. Mimic the analysis for elements of order 12 in U(720) in this chapter. 61. 60 63. They are both isomorphic to Z10 % Z4. 65. That U(n)2 is a subgroup follows from Exercise 15 of Supplementary Exercises for Chapters 1–4. 12 5 (n 2 1)2 shows that it is a proper subgroup. 67. 275 69. U(117) < U(9) % U(13) < Z6 % Z12, which contains (2, 0) % (0, 4). 71. Consider U(49). 73. Consider U(65). 75. no

Supplementary Exercises for Chapters 5–8 All things are difficult before they are easy. THOMAS FULLER

1. Consider the finite and infinite cases separately. In the finite case, note that |H| 5 |f(H)|. Now use Theorem 4.3. For the infinite case, use Exercise 2 in Chapter 6. 3. Observe that f(x21y21xy) 5 (f(x))21(f(y))21f(x)f(y), so f carries the generators of G9 to the generators of G9. 5. All nonidentity elements of G and H have order 3. G ] H.

A12

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7. Certainly the set HK has |H||K| symbols. However, not all symbols need represent distinct group elements. That is, we may have hk 5 h9k9 although h 2 h9 and k 2 k9. We must determine the extent to which this happens. For every t in H > K, hk 5 (ht)(t21k), so each group element in HK is represented by at least |H > K| products in HK. But hk 5 h9k9 implies t 5 h21h9 5 k(k9)21 [ H > K, so that h9 5 ht and k9 5 t21k. Thus each element in HK is represented by exactly |H > K| products. So, |HK| 5 |H| |K|/|H > K|. 9. U(n), where n 5 4, 8, 3, 6, 12, 24. 3 2 11. Hint: 3 1 2i 5 "13 a 1 ib "13 "13 13. Suppose f: Q → R is an isomorphism. Let f(1) 5 x0. Show that f(a/b) 5 (a/b)x0 for all integers a, b with b 2 0. 15. In Q, the equation 2x 5 a has a solution for all a. The corresponding equation x2 5 b in Q1 does not have a solution for all b. 17. Suppose x p22 5 1. Since |U(p)| 5 p 2 1, we have that x p21 5 1 for all x [ U(p). So, by cancellation, x 5 1. 19. 3 % 4 21. Z18, Z2 % Z3 % Z3, D9, D3 % Z3. 23. Say a 5 a1a2 . . . an and b 5 b1 . . . bm, where the a’s and b’s are cycles. Then ab21 5 a1a2 . . . a b 21 . . . b 21 is a finite number of cycles. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55.

57.

59.

n m

1

Count elements of order 2. Count elements of order 2. x 5 fa(x) 5 axa21, so that xa 5 ax. Conversely, if G is Abelian, fa is the identity. U50(450) (4, 10) Count elements of order 2. 20; (8, 7, (3251)) Let H 5 5x [ Zp2 % Zp2| xp 5 (0, 0) 6 . Then |H| 5 p2 and every nonidentity element of (Zp2 % Zp2 )>H has order p. (12)(34)(56789) 1260 b 5 (17395)(286) Say the points in H lie on the line y 5 mx. Then (a, b) 1 H 5 {(a 1 x, b 1 mx) | x [ R}. This set is the line y 2 b 5 m(x 2 a). aH 5 bH implies a21b [ H. So (a21b)21 5 b21a [ H. Thus, Hb21a 5 H or Hb21 5 Ha21. These steps are reversible. p2 2 1 p(p 1 1) By Theorem 8.3, U(pq) ; U(p) % U(q), so an element xn in U(pq) corresponds to an element (xn1, xn2 ) [ U(p) % U(q) . It follows from Corollary 4 of Theorem 7.1 that (xn1, xn2 ) 5 (1, 1) , the identity of U(p) % U(q) . First observe that (n, n21, . . . 2,1)(12)(123 . . . n) 5 (1n). Also, (1n)(123 . . . n) 5 (123 . . . n 2 1). So, by induction, (12) and (123 . . . n) generate Sn21. This means that every 2-cycle not involving n can be generated. Now note that (1k)(1n)(1k) 5 (kn), so all 2-cycles are generated. Let b have order 2. In disjoint cycle form, b is a product of transpositions, so there must be some i missing from this product. Thus, b(i) 5 i. Pick j such that b( j) 2 j. Since s is an n-cycle, some power of s, say st, takes i to j. If b commutes with s, it commutes with st as well. Then (stb)(i) 5 st(b(i)) 5 st(i) 5 j, whereas (bst)(i) 5 b(st(i)) 5 b( j) 2 j. This proves that stb 2 bst.

A13

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Chapter 9 There’s a mighty big difference between good, sound reasons and reasons that sound good. BURTON HILLIS

1. no 3. Say i , j and let h [ Hi > Hj. Then h [ H1H2 . . . Hi . . . Hj21 > Hj 5 {e}. 5. Recall that if A and B are matrices, then det (ABA21) 5 (det A)(det B)(det A)21. 7. Let x [ G. If x [ H, then xH 5 H 5 Hx. If x o H, then xH is the set of elements in G, not in H. But Hx is also the set of elements in G, not in H. 9. G/H < Z4 G/K < Z2 % Z2 11. No, look at D3. 13. This follows directly from (ab)h 5 a(bh) for all h [ H. 15. 2 17. H 5 {0 1 20, 4 1 20, 8 1 20, 12 1 20, 16 1 20}. G/H 5 {0 1 20 1 H, 1 1 20 1 H, 2 1 20 1 H, 3 1 20 1 H}. 19. 40/10 5 4 21. By Theorem 9.5, the group has an element a of order 3 and an element b of order 11. Then |ab| 5 33. 23. `; no, (6, 3) 1 (4, 2) has order 2. 25. Z8 27. yes; no 29. Mimic the argument given in Example 13 in this chapter. 31. Certainly, every nonzero real number is of the form 6r, where r is a positive real number. Real numbers commute, and R1 > {1, 21} 5 {1}. 33. No. If G 5 H 3 K, then |g| 5 lcm(|h|, |k|) provided that |h| and |k| are finite. If |h| or |k| is infinite, so is |g|. 35. For the first question, note that 3 > 6 5 {1} and 36 > 10 5 {1}. For the second question, observe that 12 5 32162. 37. Say |g| 5 n. Then (gH)n 5 gnH 5 eH 5 H. Now use Corollary 2 to Theorem 4.1. 39. Let x [ C(H), g [ G, and h [ H. We must show that gxg21h 5 hgxg21. Note that in the expression (gxg21)h(gxg21)21 5 gxg21hgx21g21 the terms x and x21 cancel since g21hg [ H and x commutes with every element of H. Then we have (gxg21)h(gxg21)21 5 gxg21hgx21g21 5 gg21hgg21 5 h. So, gxg21 [ C(H). 41. Take G 5 Z6, H 5 {0, 3}, a 5 1, and b 5 4. 43. Use Lagrange’s Theorem and Exercise 7 of this chapter. 45. Since N # NH # G, we have |G : N| 5 |G : NH| |NH : N|. Thus, |G : H| 5 1 or |NH : N| 5 1. It follows that G 5 NH or NH 5 N. 47. Use the “G/Z Theorem.” 49. If H is normal in G, then xNhN(xN)21 5 xhx21N [ H/N, so H/N is normal in G/N. Now assume H/N is normal in G/N. Then xhx21N 5 xNhN(xN)21 [ H/N. Thus, xhx21N 5 h9N for h9 [ H. So, xhx21 5 h9n for some n [ N. 51. Say H has index n. Then (R*)n 5 {x n | x [ R*} # H. If n is odd, then (R*)n 5 R*; if n is even, then (R*)n 5 R1. So, H 5 R* or H 5 R1. 53. Use Exercise 7 and observe that VK 2 KV. 55. Suppose n1h1 and n2h2 [ NH. Then n1h1n2h2 5 n1n9h1h2 [ NH. Also (n1h1)21 5 h121n121 5 nh121 [ NH. 57. Let N 5 a, H 5 ak, and x [ G. Then, x(ak)mx21 5 (xamx21)k 5 (ar)k 5 (ak)r [ H.

A14

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59. gcd(|x|, |G/H|) 5 1 implies gcd(|xH|, |G/H|) 5 1. But |xH| divides |G/H|. Thus |xH| 5 1 and therefore xH 5 H. 61. Note that G/N is a group and use Corollary 4 of Theorem 7.1. 63. Use Theorems 9.4 and 9.3. 65. Say |gH| 5 n. Then |g| 5 nt (by Exercise 37) and |gt| 5 n. For the second part, consider Z/k. 67. It is not a group table. No, because _ is not normal in D4. 69. Use Theorem 9.3 and Theorem 7.2. 71. By Exercise 70, A5 would have an element of the form (ab)(cd) that commutes with every element of A5. Try (abc). 73. To see that H is normal, observe that xgpx21 5 (xgx21 ) p. To verify the second part, note that (gH) p 5 gpH 5 H . 75. Since H has index 2 in G, is it a normal subgroup of G and |G/H| 5 2. It follows that for every a in G we have (aH) 2 5 H. If a is an element of G of order 2n 1 1, then H 5 a2n11H 5 ( (aH) 2 ) naH 5 aH. Thus, a is in H.

Chapter 10 It’s always helpful to learn from your mistakes because then your mistakes seem worthwhile. GARRY MARSHALL

1. 3. 5. 7. 9. 11. 13. 15. 17.

19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41.

Note that det(AB) 5 (det A)(det B). Note that ( f 1 g)9 5 f9 1 g9. Observe that (xy)r 5 xryr. Odd values of r yield an isomorphism. (sf)(g1g2) 5 s(f(g1g2)) 5 s(f(g1)f(g2)) 5 s(f(g1))s(f(g2)) 5 (sf)(g1)(sf)(g2). Ker f is a normal subgroup of Ker sf. |H|/|K| 5 [Ker sf:Ker f] f((g, h) (g9, h9)) 5 f((gg9, hh9)) 5 gg9 5 f((g, h))f((g9, h9)). The kernel is {(e, h)|h [ H}. Consider f: Z % Z → Za % Zb given by f((x, y)) 5 (x mod a, y mod b) and use Theorem 10.3. (a, b) → b is a homomorphism from A % B onto B with kernel A % {e}. 3, 13, 23 Suppose f is such a homomorphism. By Theorem 10.3, Ker f 5 (8, 1), (0, 1) or (8, 0). In these cases, (1, 0) 1 Ker f has order either 16 or 8. So, (Z16 % Z2) / Ker f is not isomorphic to Z4 % Z4. Since |Ker f| is not 1 and divides 17, f is the trivial map. 5 a. The possible images are isomorphic to Z1, Z2, Z3, Z4, Z6, and Z12. b. 1 < Z36, 2 < Z18, 3 < Z12, 4 < Z9, 6 < Z6, and 12 < Z3. 4 onto; 10 to For each k with 0 # k # n 2 1, the mapping 1 → k determines a homomorphism. Use Theorem 10.3 and properties 5, 7, and 8 of Theorem 10.2. f21(7) 5 7 Ker f 5 {7, 17} 11 Ker f f((a, b) 1 (c, d )) 5 f((a 1 c, b 1 d)) 5 (a 1 c) 2 (b 1 d) 5 a 2 b 1 c 2 d 5 f((a, b)) 1 f((c, d)). Ker f 5 {(a, a) | a [ Z}. f21(3) 5 {(a 1 3, a) | a [ Z}. f(xy) 5 (xy)6 5 x6y6 5 f(x)f(y). Ker f 5 cos 60° 1 i sin 60°. Show that the mapping from K to KN/N given by k → kN is an onto homomorphism with kernel K > N. For each divisor d of k there is a unique subgroup of Zk of order d, and this subgroup is generated by f(d) elements. A homomorphism from Zn to a subgroup of Zk must carry 1 to a generator of the subgroup. Furthermore, the order of the image of 1 must divide n, so we need consider only those divisors d of k that also divide n.

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43. 45. 47. 49. 51. 53. 55. 57. 59. 61.

A15

D4, {e}, Z2, Z2 % Z2 It is divisible by 10. It is infinite. Z Let γ be the natural homomorphism from G onto G/N. Let H be a subgroup of G/N and let γ21(H ) 5 H. Then H is a subgroup of G and H/N 5 γ (H) 5 γ (γ21(H )) 5 H . The mapping g → fg is a homomorphism with kernel Z(G). ( f 1 g)(3) 5 f(3) 1 g(3). The kernel is the set of elements in Z[x] whose graphs pass through the point (3, 0). Let g belong to G. Since f(g) belongs to Z2 % Z2 5 1, 0 x 0, 1 x 1, 1, it follows that G 5 f21(1, 0) x f21(0, 1) x f21(1, 1). Moreover, each of these three subgroups is proper. Use Exercise 54 in Chapter 9 and Exercise 39 above to prove the first assertion. To verify that G/(H > K) is not cyclic, observe that it has two subgroups of order 2. Mimic Example 16. Suppose that H is a proper subgroup of G that is not properly contained in a proper subgroup of G. Then G/H has no nontrivial, proper subgroup. It follows from Exercise 24 in Chapter 7 that G/H is isomorphic to Zp for some prime p. But then for every coset g 1 H we have p(g 1 H) 5 H so that pg [ H for all g [ G. But then G 5 pG # H. Both Q and R satisfy the hypothesis.

Chapter 11 Think before you think! STANISLAW J. LEC, Unkempt Thoughts

1. n 5 4 Z4, Z2 % Z2 3. n 5 36 Z9 % Z4, Z3 % Z3 % Z4, Z9 % Z2 % Z2, Z3 % Z3 % Z2 % Z2 5. The only Abelian groups of order 45 are Z45 and Z3 % Z3 % Z5. In the first group, |3| 5 15; in the second one, |(1, 1, 1)| 5 15. Z3 % Z3 % Z5 does not have an element of order 9. 7. Z9 % Z3 % Z4; Z9 % Z3 % Z2 % Z2 9. Z4 % Z2 % Z3 % Z5 11. By the Fundamental Theorem, any finite Abelian group G is isomorphic to some direct product of cyclic groups of prime-power order. Now go across the direct product and, for each distinct prime you have, pick off the largest factor of the prime-power. Next, combine all of these into one factor (you can do this, since the subscripts are relatively prime). Let us call the order of this new factor n1. Now repeat this process with the remaining original factors and call the order of the resulting factor n2. Then n2 divides n1, since each prime-power divisor of n2 is also a primepower divisor of n1. Continue in this fashion. Example: If G < Z27 % Z3 % Z125 % Z25 % Z4 % Z2 % Z2, then G < Z27 ? 125 ? 4 % Z3 ? 25 ? 2 % Z2. Now note that 2 divides 3 ? 25 ? 2 and 3 ? 25 ? 2 divides 27 ? 125 ? 4. 13. Z2 % Z2 15. a. 1 b. 1 c. 1 d. 1 e. 1 f. There is a unique Abelian group of order n if and only if n is not divisible by the square of any prime. 17. Z2 % Z2 19. Z3 % Z3 21. n is square-free (no prime factor of n occurs more than once).

A16

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23. Among the first 11 elements in the table, there are nine elements of order 4. None of the other isomorphism classes has this many. 25. Z4 % Z2 % Z2; one internal direct product is 7 3 101 3 199. 27. 3; 6; 12 29. Z4 % Z4 31. Use Theorems 11.1, 8.1, and 4.3. 33. |aK| 5 |a||K|/|a > K| 5 |a||K| 5 |a||K |p 5 |G|p 5 |G|. 35. By the Fundamental Theorem of Finite Abelian Groups, it suffices to show that every group of the form Zp n1 % Zp n 2 % . . . % Zp nk is a subgroup of a U-group. Consider first a group of the form 1 2 k Zp n1 % Zp n 2 ( p1 and p2 need not be distinct). By Dirichlet’s Theorem, for some s and t there are 1 2 distinct primes q and r such that q 5 tp1n1 1 1 and r 5 sp2n2 1 1. Then U(qr) 5 U(q) % U(r) < Ztp n1 % Zsp n2, and this latter group contains a subgroup isomorphic to Zp n1 % Zp n 2. The general 1 2 1 2 case follows in the same way. 37. Look at D4.

Supplementary Exercises for Chapters 9–11 You cannot have success without the failures. H. G. HASLER, The Observer

1. Say aH 5 Hb. Then a 5 hb for some h in H. Then Ha 5 Hhb 5 Hb 5 aH. 3. Suppose diag(G) is normal. Then (e, a)(b, b)(e, a)21 5 (b, aba21) [ diag(G). Thus b 5 aba21. If G is Abelian, (g, h)(b, b) ? (g, h)21 5 (gbg21, hbh21) 5 (b, b). The index of diag(G) is |G|. 5. Let a [ Aut(G) and fa [ Inn(G). Then (afaa21)(x) 5 (afa)(a21(x)) 5 a(aa21(x)a21) 5 a(a)x(a(a))21 5 fa(a)(x). 7. R* (See Example 2 in Chapter 10.) 9. a. 1 0 b Z(H) 5 • £ 0 0 b. The mapping

11. 13. 15. 17. 19.

21.

1 0

0 § † b [ Q¶ 1

1 0 b £ 0 1 0§ →b 0 0 1 is an isomorphism. c. The mapping 1 a b £ 0 1 c § → (a, c) 0 0 1 is a homomorphism with Z(H) as the kernel. d. The proofs are valid with R and Zp. b(a/b 1 Z) 5 a 1 Z 5 Z Use Exercise 5 of the Supplementary Exercises for Chapters 1– 4. Such a set is possible only when n is prime. For the first example, consider Dp, where p is a prime. For the second example, try D4. Observe that hkh21k21 5 (hkh21)k21 [ K and hkh21k21 5 h(kh21k21) [ H. Use Theorem 7.3 and Exercise 7 of Chapter 9. First observe that f((4, 0, 0)) 5 f(4(1, 0, 0)) 5 4f(1, 0, 0) 5 (0, 0), so that Ker f 5 {(0, 0, 0), (4, 0, 0)}. But then (Z8 % Z2 % Z2)/Ker f has more than three elements of order 2, whereas Z4 % Z4 has only three. Use Theorem 7.2 together with the fact that S4 has no element of order 6.

Selected Answers

A17

23. The number is m in all cases. 25. The mapping g → gn is a homomorphism from G onto Gn with kernel Gn. 27. Let |H| 5 p. Exercise 7 of Supplementary Exercises for Chapters 5–8 shows that H is the only subgroup of order p. But xHx21 is also a subgroup of order p. So, xHx21 5 H. 29. Say a and b are integers and a/b 1 Z has order n in Q/Z. Then na/b 5 m for some integer m. Thus, a/b 1 Z 5 m/n 1 Z 5 m(1/n 1 Z) [ 1/n 1 Z. 31. If (1, 0) → a and (0, 1) → b, then (x, 0) → ax and (0, y) → by. 33. First note that by Exercise 11, every element in Q/Z has finite order. For each positive integer n, let Bn denote the set of elements of order n and suppose that f is an isomorphism from Q/Z to itself. Then, by property 5 of Theorem 6.2, f(Bn) # Bn. By Exercise 29 we know that Bn is finite, and since f preserves orders and is one-to-one, we must have f(Bn) 5 Bn. Since it follows from Exercise 11 and Exercise 29 that Q/Z 5 < Bn, where the union is taken over all positive integers n, we have f(Q/Z) 5 Q/Z. 35. If the group is not Abelian, for any element a not in the center, the inner automorphism induced by a is not the identity; if the group is Abelian and contains an element a with |a| . 2, then x → x21 works; if every nonidentity element has order 2, then G is isomorphic to a group of the form Z2 % Z2 % . . . % Z2 . In this case, the mapping that takes (a1, a2, a3, . . . , ak) to (a2, a1, a3, . . . , ak) is not the identity. 37. G/H is isomorphic to Z2 % Z2. G/H is not isomorphic to a subgroup of G since G has only one element of order 2. 1 x 1 y 1 x1y 39. Observe that c dc d 5 c d, 0 1 0 1 0 1 1 x 21 1 2x d 5 c d, 0 1 0 1 which is in H. Thus, H is a subgroup of G. so H is closed. Also, c

1 a 1 x 1 a 21 1 dc dc d 5 c 0 b 0 1 0 b 0 we have that H is normal in G. Since c

a 1 dc b 0

x 1 dc 1 0

2ab21 1 21 d 5 c b 0

b21x d belongs to H, 1

41. Let g belong to G. Since gKg21 5 K , conjugation is an automorphism of K. Thus gHg21 5 H .

Chapter 12 Mistakes are the postals of discovery. JAMES JOYCE

1. For any n . 1, the ring M2(Zn ) of 2 3 2 matrices with entries from Zn is a finite noncommutative ring. The set M2(2Z ) of 2 3 2 matrices with even integer entries is an infinite noncommutative ring that does not have a unity. 3. In R, consider 5n"2 0 n [ Z 6. 5. The proofs given for a group apply to a ring as well. 7. In Zp, nonzero elements have multiplicative inverses. Use them. 9. If a and b belong to the intersection, then they belong to each member of the intersection. Thus a 2 b and ab belong to each member of the intersection. So, a 2 b and ab belong to the intersection. 11. Part 3: 0 5 0(2b) 5 (a 1 (2a))(2b) 5 a(2b) 1 (2a)(2b) 5 2(ab) 1 (2a)(2b). So, ab 5 (2a)(2b). Part 4: a(b 2 c) 5 a(b 1 (2c)) 5 ab 1 a(2c) 5 ab 1 (2(ac)) 5 ab 2 ac. Part 5: Use part 2. Part 6: Use part 3.

A18

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13. Hint: Z is a cyclic group under addition, and every subgroup of a cyclic group is cyclic. 15. For positive m and n, observe that (m ? a)(n ? b) 5 (a 1 a 1 . . . 1 a)(b 1 b 1 . . . 1 b) 5 (ab 1 ab 1 . . . 1 ab), where the last term has mn summands. Similar arguments apply in the remaining cases. 17. From Exercise 15, we have (n ? a)(m ? a) 5 (nm) ? a2 5 (mn) ? a2 5 (m ? a)(n ? a). 19. Let a, b belong to the center. Then (a 2 b)x 5 ax 2 bx 5 xa 2 xb 5 x(a 2 b). Also, (ab)x 5 a(bx) 5 a(xb) 5 (ax)b 5 (xa)b 5 x(ab). 21. (x1, . . . , xn)(a1, . . . , an) 5 (x1, . . . , xn) for all xi in Ri if and only if xiai 5 xi for all xi in Ri and i 5 1, . . . , n. 23. {1, 21, i, 2i} 25. f (x) 5 1 and g(x) 5 21. 27. If a is a unit, then b 5 a(a21b). 29. Consider a21 2 a22b. 31. Try the ring M2(Z). 33. Note that 2x 5 (2x)3 5 8x3 5 8x. 35. For Z6 use n 5 3. For Z10 use n 5 5. Say m 5 p2t where p is a prime. Then (pt)n 5 0 in Zm since m divides (pt)n. 37. Every subgroup of Zn is closed under multiplication. 39. ara 2 asa 5 a(r 2 s)a. (ara)(asa) 5 ara2sa 5 arsa. a1a 5 a2 5 1, so 1 [ S. 41. The subring test is satisfied. 43. Look at (1, 0, 1) and (0, 1, 1). 45. Observe that n ? 1 2 m ? 1 5 (n 2 m) ? 1. Also, (n ? 1)(m ? 1) 5 (nm) ? ((1)(1)) 5 (nm) ? 1. 47. {m/2n | m [ Z, n [ Z1} 49. (a 1 b)(a 2 b) 5 a2 1 ba 2 ab 2 b2 5 a2 2 b2 if and only if ba 2 ab 5 0. 51. Z2 % Z2; Z2 % Z2 % . . . (infinitely many copies).

Chapter 13 Work now or wince later. B. C. FORBES, Epigrams

1. The verifications for Examples 1–6 follow from elementary properties of real and complex numbers. For Example 7, note that c

1 0

0 0 dc 0 0

0 0 d 5 c 1 0

0 . d 0

For Example 8, note that (1, 0)(0, 1) 5 (0, 0). 3. Let ab 5 0 and a 2 0. Then ab 5 a ? 0, so b 5 0. 5. Let k [ Zn. If gcd(k, n) 5 1, then k is a unit. If gcd(k, n) 5 d . 1, write k 5 sd. Then k(n/d) 5 sd(n/d) 5 sn 5 0. 7. Let s [ R, s 2 0. Consider the set S 5 {sr | r [ R}. If S 5 R, then sr 5 1 (the unity) for some r. If S 2 R, then there are distinct r1 and r2 such that sr1 5 sr2. In this case, s(r1 2 r2) 5 0. To see what happens when the “finite” condition is dropped, consider Z. 9. (a1 1 b1"d) 2 (a2 1 b2"d) 5 (a1 2 a2) 1 (b1 2 b2)"d; (a1 1 b1"d)(a2 1 b "d) 5 2 (a1a2 1 b1b2d) 1 (a1b2 1 a2b1)"d. Thus the set is a ring. Since Z["d] is a subring of the ring of complex numbers, it has no zero-divisors. 11. The even integers. 13. (1 2 a)(1 1 a 1 a2 1 . . . 1 an21 ) 5 1 1 a 1 a2 1 . . . 1 an21 2 a 2 a2 2 . . . 2 an 5 1 2 an 5 1 2 0 5 1. 15. Suppose a 2 0 and an 5 0 (where we take n to be as small as possible). Then a ? 0 5 0 5 an 5 a ? an21, so by cancellation, an21 5 0.

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17. 19. 21. 23. 25. 27. 29.

31. 33. 35.

37. 39. 41.

43. 45.

47. 49. 51. 53. 55. 57. 59. 61. 63. 65.

A19

If a2 5 a and b2 5 b, then (ab) 2 5 a2b2 5 ab. The other cases are similar. Suppose that a is an idempotent and an 5 0. By the previous exercise, a 5 0. (3 1 4i)2 5 3 1 4i. a2 5 a implies a(a 2 1) 5 0. So if a is a unit, a 2 1 5 0 and a 5 1. See Theorems 3.1 and 12.3. Note that ab 5 1 implies aba 5 a. Thus 0 5 aba 2 a 5 a(ba 2 1). So, ba 2 1 5 0. A subdomain of an integral domain D is a subset of D that is an integral domain under the operations of D. To show that P is a subdomain, show that it is a subring and contains 1. Every subdomain contains 1 and is closed under addition and subtraction, so every subdomain contains P. |P| 5 char D when char D is prime and |P| is infinite when char D is 0. Use Theorems 13.3, 13.4, and 7.1 (Lagrange’s Theorem). By Exercise 32, 1 is the only element of an integeral domain if and only if 1 5 21. This is true only for fields of characteristic 2. a. Since a3 5 b3, a6 5 b6. Then a 5 b because we can cancel a5 from both sides (since a5 5 b5). b. Use the fact that there exist integers s and t such that 1 5 sn 1 tm, but remember that you cannot use negative exponents in a ring. (1 2 a)2 5 1 2 2a 1 a2 5 1 2 2a 1 a 5 1 2 a. Z8 Let S 5 {a1, a2, . . . , an} be the nonzero elements of the ring. First show that S 5 {a1a1, a1a2, . . . , a1an}. Thus, a1 5 a1ai for some i. Then ai is the unity, for if ak is any element of S, we have a1ak 5 a1ai a k, so that a1(ak 2 aiak) 5 0. Say |x| 5 n and |y| 5 m with n , m. Consider (nx)y 5 x(ny). a. Use the Binomial Theorem. b. Use part a and induction. c. Consider {0, 3, 6, 9} under addition and multiplication modulo 12. Use Theorems 13.4 and 9.5 and Exercise 43. a b 0 0 n c d5 c d for all members of M2(R) if and only if na 5 0 for all a in R. c d 0 0 Use Exercise 50. a. 2 b. 2, 3 c. 2, 3, 6, 11 d. 2, 3, 9, 10 2 See Example 10. Use Exercise 25 and part a of Exercise 45. Choose a 2 0 and a 2 1 and consider 1 1 a. f(x) 5 f(x . 1) 5 f(x) . f(1) so f(1) 5 1. Also, 1 5 f(1) 5 f(xx21) 5 f(x) f(x21). Since a field of order 27 has characteristic 3, we have 3a 5 0 for all a. From this, we have 6a 5 0 and 5a 5 2a.

Chapter 14 Not one student in a thousand breaks down from overwork. WILLIAM ALLAN NEILSON

1. Let r1a and r2a belong to a. Then r1a 2 r2a 5 (r1 2 r2)a [ a. If r [ R and r1a [ a, then r(r1a) 5 (rr1)a [ a. 3. Clearly, I is not empty. Now observe that (r1a1 1 . . . 1 rnan) 2 (s1a1 1 . . . 1 snan) 5 (r1 2 s1)a1 1 . . . 1 (rn 2 sn)an [ I. Also, if r [ R, then r(r1a1 1 . . . 1 rnan) 5 (rr1)a1 1 . . . 1 (rrn)an [ I. That I # J follows from closure under addition and multiplication by elements from R. 5. Let a 1 bi, c 1 di [ S. Then (a 1 bi) 2 (c 1 di) 5 a 2 c 1 (b 2 d)i and b 2 d is even. Also, (a 1 bi)(c 1 di) 5 ac 2 bd 1 (ad 1 cb)i and ad 1 cb is even. Finally, (1 1 2i)(1 1 i) 5 21 1 3i o S.

A20

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7. Since ar1 2 ar2 5 a(r1 2 r2) and (ar1)r 5 a(r1r), 4R 5 {. . . , 216, 28, 0, 8, 16, . . .}. 9. If n is prime, use Euclid’s Lemma (Chapter 0). If n is not prime, say n 5 st where s , n and t , n, then st belongs to nZ but s and t do not. 11. a. a 5 1 b. a 5 3 c. a 5 gcd(m, n) 13. a. a 5 12 b. a 5 48. To see this, note that every element of 68 has the form 6t18k1 1 6t28k2 1 . . . 1 6tn8kn 5 48s [ 48. So, 68 # 48. Also, since 48 [ 68, we have 48 # 68. c. a 5 mn 15. Let r [ R. Then r 5 1r [ A. 17. Let u [ I be a unit and let r [ R. Then r 5 r(u21u) 5 (ru21)u [ I. 19. Observe that 2 and 3 are the only nontrivial ideals of Z6, so both are maximal. More generally, Zpq , where p and q are distinct primes, has exactly two maximal ideals. 21. Clearly, I is closed under subtraction. Also, if b1, b2, b3, and b4 are even, then every entry of c

a1 a2 b1 b2 dc d is even. a3 a4 b3 b4

2q1 1 r1 2q2 1 r2 23. Use the observation that every member of R can be written in the form c d. 2q 3 1 r3 2q4 1 r4 2q1 1 r1 2q2 1 r2 r1 r2 Then note that c d 1I5 c d 1 I. 2q3 1 r3 2q4 1 r4 r3 r4 25. (br1 1 a1) 2 (br2 1 a2) 5 b(r1 2 r2) 1 (a1 2 a2) [ B; r9(br 1 a) 5 b(r9r) 1 r9a [ B. 27. Use Exercise 17. 29. Since every element of x has the form xg(x), we have x # I. If f(x) [ I, then f(x) 5 anxn 1 . . . 1 a1x 5 x(anxn21 1 . . . 1 a1) [ x. 31. Suppose f(x) 1 A 2 A. Then f(x) 1 A 5 f(0) 1 A and f(0) 2 0. Thus, 1 1 A. ( f(x) 1 A) 21 5 f(0) This shows that R/A is a field. Now use Theorem 14.4. 33. Since (3 1 i)(3 2 i) 5 10, 10 1 3 1 i 5 0 1 3 1 i. Also, i 1 3 1 i 5 23 1 3 1 i 5 7 1 3 1 i. So, Z[i]/3 1 i 5 {k 1 3 1 i | k 5 0, 1, . . . , 9}, since 1 1 3 1 i has additive order 10. 35. Use Theorems 14.3 and 14.4. 37. Since every f(x) in x, 2 has the form f(x) 5 xg(x) 1 2h(x), we have f(0) 5 2h(0), so that f(x) [ I. If f(x) [ I, then f(x) 5 anxn 1 ? ? ? 1 a1x 1 2k 5 x(anxn21 1 ? ? ? 1 a1) 1 2k [ x, 2. I is prime and maximal. Z[x]/I has two elements. 39. 3x 1 1 1 I 41. Every ideal is a subgroup. Every subgroup of a cyclic group is cyclic. 43. Use Exercise 42. 45. Say b, c [ Ann(A). Then (b 2 c)a 5 ba 2 ca 5 0 2 0 5 0. Also, (rb)a 5 r(ba) 5 r ? 0 5 0. 47. a. 3 b. 3 c. 3 49. Suppose (x 1 N(0))n 5 0 1 N(0). We must show that x [ N(0). We know that xn 1 N(0) 5 0 1 N(0), so that xn [ N(0). Then, for some m, (xn)m 5 0, and therefore x [ N(0). 51. The set Z2[x]/x2 1 x 1 1 has only four elements and each of the nonzero ones has a multiplicative inverse. For example, (x 1 x2 1 x 1 1)(x 1 1 1 x2 1 x 1 1) 5 1 1 x2 1 x 1 1. 53. x 1 2 1 x2 1 x 1 1 is not zero, but its square is. 55. If f and g [ A, then ( f 2 g)(0) 5 f (0) 2 g(0) is even and ( f ? g)(0) 5 f (0) ? g(0) is even. f(x) 5 21 [ R and g(x) 5 2 [ A, but f(x)g(x) o A. 57. Hint: Any ideal of R/I has the form A/I, where A is an ideal of R. 59. Use the fact that R/I is an integral domain to show that R/I 5 {I, 1 1 I}.

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A21

61. x , x, 2n , x, 2n21 , . . . , x, 2 63. Taking r 5 1 and s 5 0 shows that a [ I. Taking r 5 0 and s 5 1 shows that b [ I. If J is any ideal that contains a and b, then it contains I because of the closure conditions.

Supplementary Exercises for Chapters 12–14 If at first you don’t succeed, try, try, again. Then quit. There’s no use being a damn fool about it. W. C. FIELDS

1. In Z10 they are 0, 1, 5, and 6. In Z20, they are 0, 1, 5, and 16. In Z30, they are 0, 1, 6, 10, 15, 16, 21, and 25. 3. We must show that an 5 0 implies a 5 0. First show this for the case when n is a power of 2. If n is not a power of 2, say 13, for example, note that a13 5 0 implies a16 5 0. 5. Suppose A s C and B s C. Pick a [ A and b [ B such that a, b o C. But ab [ C and C is prime. 7. {0} % {0}, R % R, R % {0}, and {0} % R. The ideals of F % F are {0} % {0}, F % F, F % {0}, and {0} % F. 9. Suppose that am mod n 5 0. Since n divides am, every prime p divisor of n divides am. By Euclid’s Lemma (Chapter 0), p divides a, and since n is square-free, if follows that n divides a. 11. Suppose a1, a2 [ A but a1 o B and a2 o C. Use a1 1 a2 to derive a contradiction. 13. Clearly a contains the right-hand side. Now show that the right-hand side contains a and is an ideal. 15. Since A is an ideal, ab [ A. Since B is an ideal, ab [ B. So ab [ A > B 5 {0}. 17. 6 19. Use Exercise 4. 21. Consider x2 1 1 1 x4 1 x2. 23. Consider Z8. 25. Say char R 5 p (remember p must be prime). Then char R/A 5 the additive order of 1 1 A. But |1 1 A| divides |1| 5 p. 27. Use Theorems 13.2, 14.3, and 14.4. a b 1 1 1 0 1 0 29. Observe that A 5 e c d ` a, b [ Z2 f but c dc d 5 c d is not in A. 0 0 1 1 0 0 1 0 31. Z[i]/A has two elements. (From this it follows that A is maximal. See Theorem 14.4.) 33. A finite subset of a field is a subfield if it contains a nonzero element and is closed under addition and multiplication. 35. Observe that (a 1 bi) (a 2 bi) 5 a2 1 b2. 37. 5 39. The inverse is 2x 1 3. 41. Observe that Z5[x, y]/x, y < Z5 and use Theorem 14.4. 43. Say (a,b)n 5 (0, 0). Then an 5 0 and bn 5 0. If am 5 0 and bn 5 0, then (a, b)mn 5 ((am)n, (bn)m) 5 (0, 0). 45. If a2 5 a, then pk|a(a 2 1). Since a and a 2 1 are relatively prime, pk|a or pk|(a 2 1). So, a 5 0 or a 5 1. 47. In Z3 3"24 , (a 1 b"2 )21 5 (a 2 b"2 )/(a2 2 2b2) 5 (a 2 b"2 )/(a2 1 b2). In Z7 3"24 , (1 1 2"2 ) (1 1 5"2 ) 5 0. 49. If xn 5 0, then (rx)n 5 rnxn 5 0.

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Chapter 15 For every problem there is a solution which is simple, clean and wrong. H. L. MENCKEN

1. Part 3: f(A) is a subgroup because f is a group homomorphism. Let s [ S and f(r) 5 s. Then sf(a) 5 f(r)f(a) 5 f(ra) and f(a)s 5 f(a)f(r) 5 f(ar). Part 4: Let a and b belong to f21(B) and r belong to R. Then f(a) and f(b) are in B. So, f(a) 2 f(b) 5 f(a) 1 f(2b) 5 f(a 2 b) [ B. Thus, a 2 b [ B. Also, f(ra) 5 f(r)f(a) [ B and f(ar) 5 f(a)f(r) [ B. So, ra and ar [ f21(B). 3. We already know the mapping is an isomorphism of groups. Let F(x 1 Ker f) 5 f(x). Note that F((r 1 Ker f)(s 1 Ker f)) 5 F(rs 1 Ker f) 5 f(rs) 5 f(r)f(s) 5 F(r 1 Ker f) F(s 1 Ker f). 5. f(2 1 4) 5 f(1) 5 5, whereas f(2) 1 f(4) 5 0 1 0 5 0. 7. Observe that (x 1 y)/1 5 x/1 1 y/1 and (xy)/1 5 x/1y/1. 9. a 5 f(1) 5 f(1·1) 5 f(1)f(1) 5 aa 5 a2. 11. If a and b (b 2 0) belong to every member of the collection, then so do a 2 b and ab21. Thus, by Exercise 25 in Chapter 13, the intersection is a subfield. 13. Apply the definition. 15. Multiplication is not preserved. 17. yes 19. The set of all polynomials passing through the point (1, 0). 21. For Z6 to Z6, 1 S 0, 1 S 1, 1 S 3, and 1 S 4 each define a homomorphism. For Z20 to Z30, 1 S 0, 1 S 6, 1 S 15, and 1 S 21 each define a homomorphism. 23. The zero map and the identity map. 25. Use Exercise 24. 27. Say 1 is the unity of R. Let s 5 f(r) be any element of S. Then f(1)s 5 f(1)f(r) 5 f(1r) 5 f(r) 5 s. Similarly, sf(1) 5 s. 29. Observe that an idempotent must map to an idempotent. It follows that (a, b) → a, (a, b) → b, and (a, b) → 0 are the only ring homomorphisms. 31. Say m 5 akak21 . . . a1a0 and n 5 bkbk21 . . . b1b0. Then m 2 n 5 (ak 2 bk)10k 1 (ak21 2 bk21)10k21 1 . . . 1 (a1 2 b1)10 1 (a0 2 b0). Now use the test for divisibility by 9. 33. Use the appropriate divisibility tests. 35. Mimic Example 8. 37. Use Exercise 35. 39. Look at both sides mod 2. 41. Observe that (2 ? 1075 1 2) mod 3 5 1 and (10100 1 1) mod 3 5 2 5 21 mod 3. 43. This follows directly from Theorem 13.3 and Theorem 10.1, part 3. 45. No. The kernel must be an ideal. 47. a. Suppose ab [ f21(A). Then f(a)f(b) [ A, so that a [ f21(A) or b [ f21(A). b. Consider the natural homomorphism from R to S/A. Then use Theorems 15.3 and 14.4. 49. a. f((a, b) 1 (a9, b9)) 5 f((a 1 a9, b 1 b9)) 5 a 1 a9 5 f((a, b)) 1 f((a9, b9)) so f preserves addition. Also, f((a, b)(a9, b9)) 5 f((aa9, bb9)) 5 aa9 5 f((a, b))f((a9, b9)). b. f(a) 5 f(b) implies that (a, 0) 5 (b, 0), which implies that a 5 b. f(a 1 b) 5 (a 1 b, 0) 5 (a, 0) 1 (b, 0) 5 f(a) 1 f(b). Also, f(ab) 5 (ab, 0) 5 (a, 0)(b, 0) 5 f(a) f(b). c. Use (r, s) → (s, r). 51. Observe that x4 5 1 has two solutions in R but four in C. 53. Use Exercises 46 and 52. 55. If a / b 5 a9/ b9 and c / d 5 c9 / d9, then ab9 5 ba9 and cd9 5 dc9. So, acb9d9 5 (ab9)(cd9) 5 (ba9)(dc9) 5 bda9c9. Thus, ac / bd 5 a9c9/ b9d9 and therefore (a / b)(c / d) 5 (a9/ b9)(c9/ d9). 57. First note that any field containing Z and i must contain Q[i]. Then prove (a 1 bi)/(c 1 di) [ Q[i]. 59. The subfield of E is {ab21 | a, b [ D, b 2 0}.

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A23

61. Reflexive and symmetric properties follow from the commutativity of D. For transitivity, assume a/b ; c/d and c/d ; e/f. Then adf 5 (bc)f 5 b(cf ) 5 bde, and cancellation yields af 5 be. 63. Try ab21 → a/b. 65. The mapping a 1 bi → a 2 bi is a ring isomorphism of C. 67. Certainly the unity 1 is contained in every subfield. So, if a field has characteristic p, the subfield {0, 1, . . . , p 2 1} is contained in every subfield. If a field has characteristic 0, then {(m ? 1)(n ? 1)21 | m, n [ Z, n 2 0} is a subfield contained in every subfield. This subfield is isomorphic to Q [map (m ? 1)(n ? 1)21 to m/n]. 69. The mapping f(x) 5 (x mod m, x mod n) from Zmn to Zm { Zn is a ring isomorphism.

Chapter 16 You know my methods. Apply them! SHERLOCK HOLMES, The Hound of the Baskervilles

1. f 1 g 5 3x4 1 2x3 1 2x 1 2 f ? g 5 2x7 1 3x6 1 x5 1 2x4 1 3x2 1 2x 1 2 3. 1, 2, 4, 5 5. Write f(x) 5 (x 2 a)q(x) 1 r(x). Since deg (x 2 a) 5 1, deg r(x) 5 0 or r(x) 5 0. So r(x) is a constant. Also, f(a) 5 r(a). 7. Use Corollary 1 of Theorem 16.2. 9. Let f(x), g(x) [ R[x]. By inserting terms with the coefficient 0, we may write f(x) 5 anxn 1 . . . 1 a0 and g(x) 5 bnxn 1 . . . 1 b0. Then

f( f(x) 1 g(x)) 5 f(an 1 bn)xn 1 . . . 1 f(a0 1 b0)

5 (f(an) 1 f(bn))xn 1 . . . 1 f(a0) 1 f(b0) 5 (f(an)xn 1 . . . 1 f(a0)) 1 (f(bn)xn 1 . . . 1 f(b0)) 5 f( f(x)) 1 f(g(x)).

11. 13. 15. 17. 19.

21. 23. 25. 27. 29. 31. 33. 35.

Multiplication is done similarly. Quotient, 2x2 1 2x 1 1; remainder, 2 It is its own inverse. No. See Exercise 17. If f(x) 5 anxn 1 . . . 1 a0 and g(x) 5 bmxm 1 . . . 1 b0, then f(x) ? g(x) 5 anbmxm1n 1 . . . 1 a0b0. Let m be the multiplicity of b in q(x). Then we may write f(x) 5 (x 2 a)n (x 2 b)m q9(x), where q9(x) is in F[x] and q9(b) Z 0. This means that b is a zero of f(x) of multiplicity at least m. If b is a zero of f(x) greater than m, then b is a zero of g(x) 5 f(x)/(x 2 b)m 5 (x 2 a)nq9(x). But then 0 5 g(b) 5 (b 2 a)n q9(b), and therefore q9(b) 5 0. Use Corollary 3 of Theorem 16.2. If f(x) 2 g(x), then deg[f(x) 2 g(x)] , deg p(x). But the minimum degree of any member of p(x) is deg p(x). Start with (x 2 1/2)(x 1 1/3) and clear fractions. “Long divide” x 2 a into f(x) and induct on deg f(x). By Theorem 16.4, I 5 x 2 1. Use Corollary 2 of Theorem 15.5 and Exercise 9 in this chapter. For any a in U(p), ap21 5 1, so every member of U(p) is a zero of x p21 2 1. Now use the Factor Theorem and a degree argument. Use Exercise 34.

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37. Observe that, modulo 101, (50!)2 5 (50!)(21)(22) . . . (250) 5 (50!)(100)(99) . . . (51) 5 100! and use Exercise 34. 39. Take R 5 Z and I 5 2. 41. Hint: F[x] is a PID. So  f(x), g(x) 5 a(x) for some a(x) [ F[x]. Thus a(x) divides both f(x) and g(x). This means that a(x) is a constant. 43. Write f(x) 5 (x 2 a)g(x). Use the product rule to compute f9(x). 45. Say deg g(x) 5 m, deg h(x) 5 n, and g(x) has leading coefficient a. Let k(x) 5 g(x) 2 axm2nh(x). Then deg k(x) , deg g(x) and h(x) divides k(x) in Z[x] by induction. So, h(x) divides k(x) 1 axm2nh(x) 5 g(x) in Z[x]. 47. Consider the remainder when x43 is divided by x2 1 x 1 1. 49. Observe that every term of f(a) has the form ci ai and ci ai mod m 5 ci bi mod m. To prove the second statement, assume that there is some integer k such that f(k) 5 0. If k is even, then because k mod 2 5 0, we have by the first statement 0 5 f(k) mod 2 5 f(0) mod 2 so that f(0) is even. This shows that k is not even. If k is odd, then k mod 2 5 1, so by the first statement f(k) 5 0 is odd. This contradiction completes the proof. 51. A solution to x25 2 1 5 0 in Z37 is a solution to x25 5 1 in U(37). So, by Corollary 2 of Theorem 4.1, |x| divides 25. Moreover, we must also have that |x| divides |U(37)| 5 36.

Chapter 17 Experience enables you to recognize a mistake when you make it again. FRANKLIN P. JONES

1. Use Theorem 17.1. 3. If f(x) is not primitive, then f(x) 5 ag(x), where a is an integer greater than 1. Then a is not a unit in Z[x] and f(x) is reducible. 5. a. If f(x) 5 g(x)h(x), then af(x) 5 ag(x)h(x). b. If f(x) 5 g(x)h(x), then f(ax) 5 g(ax)h(ax). c. If f(x) 5 g(x)h(x), then f(x 1 a) 5 g(x 1 a)h(x 1 a). d. Try a 5 1. 7. Find an irreducible polynomial p(x) of degree 2 over Z5. Then Z5[x]/p(x) is a field of order 25. 9. Note that 21 is a zero. No, since 4 is not a prime. 11. Let f(x) 5 x4 1 1 and g(x) 5 f(x 1 1) 5 x4 1 4x3 1 6x2 1 4x 1 2. Then f(x) is irreducible over Q if g(x) is. Eisenstein’s Criterion shows that g(x) is irreducible over Q. To see that x4 1 1 is reducible over R, observe that x8 2 1 5 (x4 1 1)(x4 2 1) x4

so any complex zero of 1 1 is a complex zero of x8 2 1. Also note that the complex zeros of x4 1 1 must have order 8 (when considered as an element of C). Let v 5 "2 /2 1 i "2 /2. Then Example 2 in Chapter 16 tells us that the complex zeros of x4 1 1 are v, v3, v5, and v7, so x4 1 1 5 (x 2 v)(x 2 v3)(x 2 v5)(x 2 v7).

13. 15. 17. 19. 21. 23.

But we may pair these factors up as ((x 2 v) (x 2 v7))((x 2 v3)(x 2 v5)) 5 (x2 2 "2 x 1 1) (x2 1 "2 x 11) to factor using reals (see DeMoivre’s Theorem, Example 7 in Chapter 0). (x 1 3)(x 1 5)(x 1 6) a. Consider the number of distinct expressions of the form (x 2 c)(x 2 d). b. Reduce the problem to the case considered in part a. Use Exercise 16, and imitate Example 10. Map Z3[x] onto Z3[i] by f(x) → f(i). This is a ring homomorphism with kernel  x2 1 1. x2 1 1, x2 1 x 1 2, x2 1 2x 1 2 1 has multiplicity 1, 3 has multiplicity 2.

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A25

25. We know that an(r/s)n 1 an21(r/s)n21 1 . . . 1 a0 5 0. So anr n 1 san21r n21 1 . . . 1 sna0 5 0. This shows that s | anrn and r | sna0. Now use Euclid’s Lemma and the fact that r and s are relatively prime. 27. Use induction and Corollary 2 of Theorem 17.5. 29. If there is an a in Zp such that a2 5 21, then x4 1 1 5 (x2 1 a)(x2 2 a). If there is an a in Zp such that a2 5 2, then x4 1 1 5 (x2 1 ax 1 1)(x2 2 ax 1 1). If there is an a in Zp such that a2 5 22, then x4 1 1 5 (x2 1 ax 2 1)(x2 2 ax 2 1). To show that one of these three cases must occur, consider the group homomorphism from Z*p to itself given by x → x2. Since the kernel is {1, 21}, the image H has index 2 (we may assume that p 2 2). Suppose that neither 21 nor 2 belongs to H. Then, since there is only one coset other than H, we have 21H 5 2H. Thus, H 5 (21H)(21H) 5 (21H)(2H) 5 22H, so that 22 is in H. 31. Since ( f 1 g)(a) 5 f(a) 1 g(a) and ( f ? g)(a) 5 f(a)g(a), the mapping is a homomorphism. Clearly, p(x) belongs to the kernel. By Theorem 17.5,  p(x) is a maximal ideal, so the kernel is  p(x). 33. The mapping a → a 1  p(x) is an isomorphism. 35. Although the probability of rolling any particular sum is the same with either pair of dice, the probability of rolling doubles is different (1/6 with ordinary dice, 1/9 with Sicherman dice). Thus the probability of going to jail is different. Other probabilities are also affected. For example, if in jail one cannot land on Virginia by rolling a pair of 2’s with Sicherman dice, but one is twice as likely to land on St. James with a pair of 3’s with the Sicherman dice as with ordinary dice. 37. The analysis is identical except that 0 # q, r, t, u # n. Now, just as when n 5 2, we have q 5 r 5 t 5 1, but this time 0 # u # n. However, when u . 2, P(x) 5 x(x 1 1) ? (x2 1 x 1 1)(x2 2 x 1 1)u has (2u 1 2)x2u13 as one of its terms. Since the coefficient of x2u13 represents the number of dice with the label 2u 1 3, the coefficient cannot be negative. Thus, u # 2, as before.

Chapter 18 He thinks things through very carefully, before going off half-cocked. GENERAL CARL SPAATZ, in Presidents Who Have Known Me, GEORGE E. ALLEN

1. 1. |a2 2 db2| 5 0 implies a2 5 db2. Thus a 5 0 5 b, since otherwise d 5 1 or d is divisible by the square of a prime. 2. N((a 1 b"d )(a9 1 b9"d )) 5 N(aa9 1 dbb9 1 (ab9 1 a9b)"d ) 5 |(aa9 1 dbb9)2 2 d(ab9 1 a9b)2| 5 |a2a92 1 d 2b2b92 2 da2b92 2 da92b2| 5 |a2 2 db2||a92 2 db92| 5 N(a 1 b"d ) N(a9 1 b9"d ). 3. If xy 5 1, then 1 5 N(1) 5 N(xy) 5 N(x)N(y) and N(x) 5 1 5 N(y). If N(a 1 b"d) 5 1, then 61 5 a2 2 db2 5 (a 1 b"d) (a 2 b"d) and a 1 b"d is a unit. 4. This property follows directly from properties 2 and 3. 3. Let I 5
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15. Observe that 10 5 2 ? 5 and 10 5 (2 2 "26) (2 1 "26) and mimic Example 8. A PID is a UFD. 17. Suppose 3 5 ab, where a, b [ Z[i] and neither is a unit. Then 9 5 d(3) 5 d(a)d(b), so that d(a) 5 3. But there are no integers such that a2 1 b2 5 3. Observe that 2 5 2i(1 1 i)2 and 5 5 (1 1 2i)(1 2 2i). 19. Use Exercise 1 with d 5 21. 5 and 1 1 2i; 13 and 3 1 2i; 17 and 4 1 i. 21. Mimic Example 1. 23. (21 1 "5)(1 1 "5) 5 4 5 2 ? 2. Now use Exercise 22. 25. Use the fact that x is a unit if and only if N(x) 5 1. 27. See Example 3. 29. p|(a1a2 . . . an21)an implies that p|a1a2 . . . an21 or p|an. Thus, by induction, p divides some ai. 31. Use Exercise 10 and Theorem 14.4. 33. Suppose R satisfies the ascending chain condition and there is an ideal I of R that is not finitely generated. Then pick a1 [ I. Since I is not finitely generated, a1 is a proper subset of I, so we may choose a2 [ I but a2 o a1. As before, a1, a2 is proper, so we may choose a3 [ I but a3 o a1, a2. Continuing in this fashion, we obtain a chain of infinite length a1 , a1, a2 , a1, a2, a3 , . . .. Now suppose every ideal of R is finitely generated and there is a chain I1 , I2 , I3 , . . .. Let I 5
Supplementary Exercises for Chapters 15–18 Errors, like straws, upon the surface flow; He who would search for pearls must dive below. JOHN DRYDEN

1. Use Theorem 15.3, Supplementary Exercise 8 for Chapters 12–14, Theorem 14.4, and Example 13 in Chapter 14. 3. To show the isomorphism, use the First Isomorphism Theorem. 5. Use the First Isomorphism Theorem. 7. Consider the obvious homomorphism from Z[x] onto Z2[x]. Then use the First Isomorphism Theorem and Theorem 14.3. 9. As in Example 7 in Chapter 6, the mapping is onto, is one-to-one, and preserves multiplication. Also, a(x 1 y)a21 5 axa21 1 aya21, so that it preserves addition as well. 11. Z[i]/2 1 i 5 {0 1 2 1 i, 1 1 2 1 i, 2 1 2 1 i, 3 1 2 1 i, 4 1 2 1 i}. Note that 5 1 2 1 i 5 (2 1 i)(2 2 i) 1 2 1 i 5 0 1 2 1 i.

13. Observe that (3 1 2"2) (3 2 2"2) 5 1 . 15. In Zn we are given (k 1 1)2 5 k 1 1. So, k2 1 2k 1 1 5 k 1 1 or k2 5 2k 5 n 2 k. Also, (n 2 k)2 5 n2 2 2nk 1 k2 5 k2, so (n 2 k)2 5 n 2 k.

Selected Answers

17. 19. 21. 23. 25. 27. 29. 31. 33.

A27

Observe that for any integer a, a2 mod 4 5 0 or 1. Use the Mod 2 Irreducibility Test. Use Theorem 14.4. The factor ring has two elements. Use Theorem 14.4. Say a/b, c/d [ R. Then (ad 2 bc)/(bd) and ac/(bd) [ R by Euclid’s Lemma. The field of quotients is Q. Z[i]/3 is a field and Z3 % Z3 is not. Consider the mapping from R[x] to (R/I)[x] given by anxn 1 . . . 1 a0 → (an 1 I)x n 1 . . . 1 (a0 1 I ). Let I 5 2[x]. Then Z8[x]/I is isomorphic to Z2[x]. x, 3.

Chapter 19 When I was young I observed that nine out of every ten things I did were failures, so I did ten times more work. GEORGE BERNARD SHAW

1.

Rn

has basis {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, 0, . . . , 1)}; M2(Q) has basis ec

1 0 0 1 0 0 0 0 d, c d, c d, c df; 0 0 0 0 1 0 0 1

Zp[x] has basis {1, x, x2, . . .}; C has basis {1, i}. 3. (a2x2 1 a1x 1 a0) 1 (a29x2 1 a19x 1 a09) 5 (a2 1 a29)x2 1 (a1 1 a19)x 1 (a0 1 a09) and a(a2x2 1 a1x 1 a0) 5 aa2x2 1 aa1x 1 aa0. A basis is {1, x, x2}. Yes. 5. Linearly dependent, since 23(2, 21, 0) 2 (1, 2, 5) 1 (7, 21, 5) 5 (0, 0, 0). 7. Suppose au 1 b(u 1 v) 1 c(u 1 v 1 w) 5 0. Then (a 1 b 1 c)u 1 (b 1 c)v 1 cw 5 0. Since {u, v, w} are linearly independent, we obtain c 5 0, b 1 c 5 0, and a 1 b 1 c 5 0. So, a 5 b 5 c 5 0. 9. If the set is linearly independent, it is a basis. If not, then delete one of the vectors that is a linear combination of the others (see Exercise 8). This new set still spans V. Repeat this process until you obtain a linearly independent subset. Since the set is finite, we will eventually obtain a linearly independent set that still spans V. 11. Let u1, u2, u3 be a basis for U and w1, w2, w3 be a basis for W. Use the fact that u1, u2, u3, w1, w2, w3 are linearly dependent over F. In general, if dim U 1 dim W . dim V, then U > W 2 {0}. 13. no 15. yes; 2 a a1b a9 a9 1 b9 a 1 a9 a 1 b 1 a9 1 b9 d 1 c d5c d and 17. c a1b b a9 1 b9 b9 a 1 b 1 a9 1 b9 b 1 b9 a a1b ac ac 1 bc d 5 c d. b bc a1b ac 1 bc Suppose B is a basis. Then every member of V is some linear combination of elements of B. If a1v1 1 . . . 1 anvn 5 a19v1 1 . . . 1 an 9vn, where vi [ B, then (a1 2 a 91 )v1 1 . . . 1 (an 2 an 9)vn 5 0 and ai 2 a i 9 5 0 for all i. Conversely, if every member of V is a unique linear combination of elements of B, certainly B spans V. Also, if a1v1 1 . . . 1 anvn 5 0, then a1v1 1 . . . 1 anvn 5 0v1 1 . . . 1 0vn and ai 5 0 for all i. Since w1 5 a1u1 1 a2u2 1 . . . 1 anun and a1 2 0, we have u1 5 a121(w1 2 a2u2 2 . . . 2 anun), and therefore u1 [ w1, u2, . . . , un. Clearly, u2, . . . , un [ w1, u2, . . . , un. Hence every linear combination of u1, . . . , un is in w1, u2, . . . , un. {(1, 0, 1, 1), (0, 1, 0, 1)}. Study the proof of Theorem 19.1. cc

19.

21.

23. 25.

A28

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27. If V and W are vector spaces over F, then the mapping must preserve addition and scalar multiplication. That is, T: V → W must satisfy T(u 1 v) 5 T(u) 1 T(v) for all vectors u and v in V, and T(au) 5 aT(u) for all vectors u in V and scalars a in F. A vector space isomorphism from V to W is a one-to-one linear transformation from V onto W. 29. Suppose v and u belong to the kernel and a is a scalar. Then T(v 1 u) 5 T(v) 1 T(u) 5 0 1 0 5 0 and T(av) 5 aT(u) 5 a ? 0 5 0. 31. Let {v1, v2, . . . , vn} be a basis for V. Map a1v1 1 a2v2 1 . . . 1 anvn to (a1, a2,. . . , an).

Chapter 20 Well here’s another clue for you all. JOHN LENNON AND PAUL MCCARTNEY,

“Glass Onion,” The White Album

1. Compare with Exercise 24 in the Supplementary Exercises for Chapters 12–14. 3. Q( "23) 5. Q( "23)

7. Note that x 5 "1 1 15 implies x4 2 2x2 2 4 5 0. 9. a5 5 a2 1 a 1 1; a22 5 a2 1 a 1 1; a100 5 a2 11. The set of all expressions of the form (anpn 1 an21pn21 1 . . . 1 a0)/(bmpm 1 bm21pm21 1 . . . 1 b0), where bm 2 0. 13. x7 2 x 5 x(x6 2 1) 5 x(x3 1 1)(x3 2 1) 5 x(x 2 1)3(x 1 1)3 ; x10 2 x 5 x(x9 2 1) 5 x(x 2 1)9 (see Exercise 45 in Chapter 13). 15. Hint: Use Exercise 45 in Chapter 13. 17. a 5 4/3, b 5 2/3, c 5 5/6 19. Use the fact that 1 1 i 5 2(4 2 i) 1 5 and 4 2 i 5 5 2 (1 1 i). 21. If the zeros of f(x) are a1, a2, . . . , an, then the zeros of f(x 1 a) are a1 2 a, a2 2 a, . . . , an 2 a. Now use Exercise 20. 23. Q and Q("2) 25. Let F 5 Z3[x]/x3 1 2x 1 1 and denote the cosets x 1 x3 1 2x 1 1 by b and 2 1 x3 1 2x 1 1 by 2. Then x3 1 2x 1 1 5 (x 2 b)(x 2 b 2 1)(x 1 2b 1 1). 27. Suppose that f: Q("23) → Q("3) is an isomorphism. Since f(1) 5 1, we have f(23) 5 23. 29. 31. 33. 35. 37.

Then 23 5 f(23) 5 f( "23"23) 5 (f( "23))2. This is impossible, since f( "23) is a real number. Use long division. Use Theorem 20.5. Use Theorem 20.5. Since L is a splitting field of f(x) over F, we may write f(x) 5 (x 2 a1)(x 2 a2) . . . (x 2 an), where the coefficients of f(x) belong to F. But then these coefficients also belong to L. Since |(Z2[x]/ f(x))*| 5 31, every nonidentity is a generator.

Chapter 21 Work is the greatest thing in the world, so we should always save some of it for tomorrow. DON HERALD

1. It follows from Theorem 21.1 that if p(x) and q(x) are both monic irreducible polynomials in F[x] with p(a) 5 q(a) 5 0, then deg p(x) 5 deg q(x). If p(x) 2 q(x), then ( p 2 q)(a) 5 p(a) 2 q(a) 5 0 and deg (p(x) 2 q(x)) , deg p(x), contradicting Theorem 21.1. To prove Theorem 21.3, use the Division Algorithm (Theorem 16.2).

Selected Answers

A29

3. Note that 3Q( "2): Q4 5 n and use Theorem 21.5. 5. Use Exercise 4. 7. Suppose Q("a) 5 Q("b). If "b [ Q, then "a [ Q and we may take c 5 "a/"b. If "b o Q, then "a o Q. Write "a 5 r 1 s"b. It follows that r 5 0 and a 5 bs2. The other direction follows from Exercise 20 in Chapter 20. 9. Observe that [F(a):F] must divide [E:F]. 11. Note that [F(a, b):F] is divisible by both m 5 [F(a):F] and n 5 [F(b):F] and [F(a,b):F] # mn. 13. Note that a is a zero of x3 2 a3 over F(a3)[x]. For the second part, take F 5 Q, a 5 1; F 5 Q, a 5 3 (21 1 i"3) >2; F 5 Q, a 5 " 2. 15. Suppose E1 > E2 2 F. Then [E1:E1 > E2] ? [E1 > E2:F] 5 [E1:F] implies [E1:E1 > E2] 5 1, so that E1 5 E1 > E2. Similarly, E2 5 E1 > E2. 17. E must be an algebraic extension of R, so that E # C. But then [C:E][E:R] 5 [C:R] 5 2. 19. Let a be a zero of p(x) in some extension of F. First note [E(a):E] # [F(a):F] 5 deg p(x). Then observe that [E(a):F(a)][F(a):F] 5 [E(a):E][E:F]. This implies that deg p(x) divides [E(a):E], so deg p(x) 5 [E(a):E]. 21. Hint: If a 1 b and ab are algebraic, then so is " (a 1 b) 2 2 4ab. 23. "b2 2 4ac 25. Use the Factor Theorem. 27. Say a is a generator of F*. If char F 5 0, then the prime subfield of F is isomorphic to Q. Since Q* is not cyclic, we have that F 5 Zp(a), and it suffices to show that a is algebraic over Zp. If a [ Zp, we are done. Otherwise, 1 1 a 5 ak for some k 2 0. If k . 0, we are done. If k , 0, then a2k 1 a12k 5 1 and we are done. 29. If [K:F] 5 n, then there are elements v1, v2, . . . , vn in K that constitute a basis for K over F. The mapping a1v1 1 . . . 1 anvn → (a1, . . . , an) is a vector space isomorphism from K to F n. If K is isomorphic to Fn, then the n elements in K corresponding to (1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 1) in F n constitute a basis for K over F. 31. Observe that [F(a,b) : F(a)] 5 [F(a)(b) : F(a)] # [F(b) : F] # [F(a)(b) : F(b)] [F(b) : F] 5 [F(a)(b) : F] 5 [F(a,b) : F]. 33. Mimic Example 5. 35. Mimic Example 6. 37. Observe that K 5 F(a1, a2, . . . , an), where a1, a2, . . . , an are the zeros of the polynomial. Now use Theorem 21.5. 39. Elements of Q(p) have the form (amp m 1 am21p m21 1 . . . 1 a0)/(bnp n 1 bn21p n21 1 . . . 1 b0), where the a’s and b’s are rational numbers. So, if "2 [ Q(p) , we have an expression of the form 2(bnp n 1 bn21p n21 1 . . . 1 b0)2 5 (amp m 1 am21p m21 1 . . . 1 a0)2. Equating the lead terms n

of both sides, we have 2b 2np2n 5 a 2mp2m. But then we have m 5 n, and "2 is equal to the rational number am>bn.

41. Observe that F(a) 5 F(1 1 a21 ).

Chapter 22 Tell me tell me tell me come on tell me the answer. JOHN LENNON AND PAUL MCCARTNEY, “Helter Skelter,” The White Album

1. [GF(729):GF(9)] 5 3 [GF(64):GF(8)] 5 2 3. The lattice of subfields of GF(64) looks like Figure 21.3 with GF(2) at the bottom, GF(64) at the top, and GF(4) and GF(8) on the sides. 5. 2a 1 1

A30

Selected Answers

7. Use Theorem 22.2. 9. The only possibilities for f(x) are x3 1 x 1 1 and x3 1 x2 1 1. See Exercise 8 in Chapter 20 for the first case. See Example 2 in this chapter for the second case. 11. Use Exercise 44 in Chapter 15 and Corollary 4 of Lagrange’s Theorem (Theorem 7.1). 13. Use the fact that if g(x) is an irreducible factor of x8 2 x over Z 2 and deg g(x) 5 m, then the field Z 2[x]/g(x) has order 2m and is a subfield of GF(8). Now use Theorem 22.3. 15. Direct calculations show that given x3 1 2x 1 1 5 0, we have x2 2 1 and x13 2 1. 17. Direct calculations show that x13 5 1, whereas (2x)2 2 1 and (2x)13 2 1. Thus, 2x is a generator. 19. First observe that for any field F the set F* is a group under multiplication. Now use Theorem 22.2 and Theorem 4.3. 21. Find a quadratic irreducible polynomial p(x) over Z 3; then Z 3[x]/p(x) is a field of order 9. m

m

m

23. Let a, b [ K. Then, by Exercise 45b in Chapter 13, (a 2 b) p 5 a p 2 b p 5 a 2 b. Also, m m m (ab) p 5 a p b p 5 ab. So, K is a subfield. n21 p 25. Consider x 2 1 and use Corollary 3 of Lagrange’s Theorem (Theorem 7.1). 27. identical 29. Consider g(x) 5 x2 2 a. Note that |GF(p)[x]/g(x)| 5 p2, so that g(x) has a zero in GF(p2). Now use Theorem 22.3. 31. Use Exercise 11. 33. Since F* is a cyclic group of order 124, it has a unique element of order 2. 35. Use Exercise 45 in Chapter 13. 37. Consider the field of quotients of Zp[x]. The polynomial f(x) 5 x is not the image of any element.

Chapter 23 Why, sometimes I’ve believed as many as six impossible things before breakfast. LEWIS CARROLL

1. To construct a 1 b, first construct a. Then use a straightedge and compass to extend a to the right by marking off the length of b. To construct a 2 b, use the compass to mark off a length of b from the right end point of a line of length a. 3. Let y denote the length of the hypotenuse of the right triangle with base 1, and let x denote the length of the hypotenuse of the right triangle with base |c|. Then y2 5 1 1 d 2, y2 1 x2 5 (1 1 |c|)2, and |c|2 1 d 2 5 x2. So, 1 1 2|c| 1 |c|2 5 1 1 d 2 1 |c|2 1 d 2, which simplifies to |c| 5 d 2. 5. Use sin2 u 1 cos2 u 5 1. 7. Use cos 2u 5 2 cos2 u 21. 9. Use sin(a 2 b) 5 sin a cos b 2 cos a sin b. 11. Solving two linear equations with coefficients from F involves only the operations of F. 13. Use Theorem 17.1 and Exercise 25 in Chapter 17. 15. If so, then an angle of 40° is constructible. Now use Exercise 10. 17. This amounts to showing that "p is not constructible. But if "p is constructible, so is p. However, [Q(p) : Q] is infinite. 3 19. No, since [Q( "3 ):Q] 5 3. 3 21. No, since [Q( "p ):Q] is infinite.

Supplementary Exercises for Chapters 19–23 The things taught in colleges and schools are not an education, but the means of education. RALPH WALDO EMERSON, Journals

1. Use Theorem 20.5. 3. Suppose b is one solution of xn 5 a. Since F* is a cyclic group of order q 2 1, it has a cyclic subgroup of order n, say c. Then each member of c is a solution to the equation xn 5 1. It follows that bc is the solution set of xn 5 a.

Selected Answers

A31

5. (5a 2 1 2)/a 5 5a 1 2a21. Now observe that since a2 1 a 1 1 5 0, we know that a(2a 2 1) 5 1, and so a21 5 2a 2 1. Thus, (5a2 1 2)/a 5 22 1 3a. 7. 5 9. Since F(a) 5 F(a21), we have degree of a 5 [F(a):F] 5 [F(a21):F] 5 degree of a21. 11. If ab is a zero of cnx n 1 . . . 1 c1x 1 c0 [ F[x], then a is a zero of cnb nx n 1 . . . 1 c1bx 1 c0 [ F(b)[x]. 13. Every element of F(a) can be written in the form f(a)/g(a), where f(x), g(x) [ F[x]. If f(a)/g(a) is algebraic and not in F, then there is some h(x) [ F[x] such that h( f (a)/g(a)) 5 0. By clearing fractions and collecting like powers of a, we obtain a polynomial in a with coefficients from F equal to 0. But then a would be algebraic over F. 15. Use Corollary 2 to Theorem 22.2. 17. If the basis elements commute, then so would any combination of basis elements. However, the entire space is not commutative. 19. {x, x2, x3} 21. Use Exercise 45 in Chapter 13. 23. By Theorem 20.5, the zeros of xn 2 a are distinct, say a1, a2, . . . , an. Then bi 5 ai /a1 for i 5 1, 2, . . . , n are all the nth roots of unity.

Chapter 24 Difficulty, my brethren, is the nurse of greatness. WILLIAM CULLEN BRYANT

1. a 5 5 b implies a 5 5 a5 and b 5 ycy21 imply 21 21 21 a 5 xycy x 5 xyc(xy) . 3. Observe that T(xC(a)) 5 xax21 5 yay21 5 T(yC(a)) if and only if y21xa 5 ay21x if and only if y21x [ C(a) if and only if yC(a) 5 xC(a). This proves that T is well defined and one-to-one. T is onto by definition. 5. By way of contradiction, assume that H is the only Sylow 2-subgroup of G and that K is the only Sylow 3-subgroup of G. Then H and K are normal and Abelian (corollary to Theorem 24.5 and corollary to Theorem 24.2). So, G 5 H 3 K < H % K and, from Exercise 4 in Chapter 8, G is Abelian. 7. Use Exercise 7 in Supplementary Exercises for Chapters 5–8. 9. Use Exercise 55 in Chapter 9 and Exercise 7 of the Supplementary Exercises for Chapters 5–8. 11. By Theorem 24.5, np, the number of Sylow p-subgroups has the form 1 1 kp and np divides |m| . But if k $ 1, np does not divide |m| . Thus k 5 0. Now use the corollary to Theorem 24.5. 13. 8 15. 15 17. By Exercise 16, G has seven subgroups of order 3. 19. 10; (123), (234), (134), (345), (245) 23. 21 25. Sylow’s Third Theorem implies that the Sylow 3- and Sylow 5-subgroups are unique. Pick any x not in the union of these. Then |x| 5 15. 27. By Sylow, n17 5 1 or 35. Assume n17 5 35. Then the union of the Sylow 17-subgroups has 561 elements. By Sylow, n5 5 1. Thus, we may form a cyclic subgroup of order 85 (Exercise 55 in Chapter 9 and Theorem 24.6). But then there are 64 elements of order 85. This gives too many elements. 29. Use the “G/Z Theorem” (Theorem 9.3). 31. Let H be the Sylow 3-subgroup and suppose that the Sylow 5-subgroups are not normal. By Sylow, there must be six Sylow 5-subgroups, call them K1, . . . , K6. These subgroups have 24 elements of order 5. Also, each of the cyclic subgroups HK1, . . . , HK6 has eight generators. Thus, there are 48 elements of order 15 which results in more than 60 elements in the group. eae21;

cac21

c21bc

c21b(c21)21;

xbx21

A32

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33. Mimic the proof of Sylow’s First Theorem. 35. Pick x [ Z(G) such that |x| 5 p. If x [ H, by induction, N(H/x) . H/x, say yx [ N(H/x) but not H/x. Now show y [ N(H) but not H. If x o H, then x [ N(H), so that N(H) . H. 37. Automorphisms preserve order. 39. Since 3 divides |N (K )|, we know that N (K) has subgroup H1 of order 3. Now use the fact that H1K is cyclic group of order 15 and Exercise 25 in the Supplementary Exercises for Chapters 1– 4. 41. Normality of H implies cl(h) # H for h in H. Now observe that h [ cl(h). This is true only when H is normal. 43. The mapping from H to xHx21 given by h → xhx21 is an isomorphism. 45. Say cl(x) 5 {x, g1xg121, g2 xg221 , . . . , gkxgk21}. If x21 5 gi xgi21, then for each gj xgj21 in cl(x), we have (gjxgj21) 21 5 gj x21gj21 5 gj(gi xgi21)gj21 [ cl(x). Because |G| has odd order, gj xgj21 2 (gj xgj21)21. It follows that |cl(x)| is even. But |cl(x)| divides |G|. 47. Mimic Example 4. 49. Say cl(e) and cl(a) are the only two conjugacy classes of a group G of order n. Then cl(a) has n 2 1 elements all of the same order, say m. If m 5 2, then it follows from Exercise 35 in Chapter 2 that G is Abelian. But then cl(a) 5 {a} and so n 5 2. If m . 2, then cl(a) has at most n 2 2 elements since conjugation of a by e, a, and a2 each yields a. 51. Let H be a Sylow 5-subgroup. Since the number of Sylow 5-subgroups is 1 modulo 5 and divides 7 . 17, the only possibility is 1. So, H is normal in G. Then by the N/C Theorem (Example 15 of Chapter 10), |G>C(H)| divides both 4 and |G| . Thus C(H) 5 G. 53. This follows directly from Theorem 24.1. 55. Pr(D4) 5 5/8, Pr(S3) 5 1/2, Pr(A4) 5 1/3 57. Exactly as in the case for a group, we have for a ring R 5 {x1, x2, . . . , xn}, Pr(R) 5 |K|/n2, where K 5 {(x, y)|xy 5 yx, x, y [ R}. Also, |K| 5 |C(x1)| 1 |C(x2)| 1 . . . 1 |C(xn)|. From Exercise 28 in the Supplementary Exercises for Chapters 12–14, we know that R/C(R) is not cyclic. Thus, |R/C(R)| $ 4 and so |C(R)| # |R|/4. So, for at least 3/4 of the elements x of R, we have |C(x)| # |R|/2. Then starting with the elements in the center and proceeding to the elements not in the center, we have |K| # |R|/4 1 (1/2)(3/4)|R| 5 (5/8)|R|.

Chapter 25 Learn to reason forward and backward on both sides of a question. THOMAS BLANDI

1. Use the 2 ? odd test. 3. Use the Index Theorem. 5. Suppose G is a simple group of order 525. Let L7 be a Sylow 7-subgroup of G. It follows from Sylow’s theorems that |N(L 7)| 5 35. Let L be a subgroup of N(L 7) of order 5. Since N(L 7) is cyclic (Theorem 24.6), N(L) $ N(L 7), so that 35 divides |N(L)|. But L is contained in a Sylow 5-subgroup (Theorem 24.4), which is Abelian (see the Corollary to Theorem 24.2). Thus, 25 divides |N(L)| as well. It follows that 175 divides |N(L)|. The Index Theorem now yields a contradiction. 7. n11 5 12. Use the N/C Theorem (Example 15 in Chapter 10) to show that there is an element of order 22; then use the Embedding Theorem and observe that A12 has no element of order 22. 9. Suppose that there is a simple group of order 396 and L11 is a Sylow 11-subgroup. Use the N/C Theorem given in Example 15 of Chapter 10 to show that C(L11) has an element of order 33 whereas A12 does not. 11. If we can find a pair of distinct Sylow 2-subgroups A and B such that |A > B| 5 8, then N(A > B) $ AB, so that N(A > B) 5 G. Now let H and K be any distinct pair of Sylow 2-subgroups. Then 16 ? 16/|H > K| 5 |HK| # 112 (Supplementary Exercise 7 for Chapters 5–8), so that

Selected Answers

15. 17.

19. 21.

23.

25. 27.

29.

31.

A33

|H > K| is at least 4. If |H > K| 5 8, we are done. So, assume |H > K| 5 4. Then N(H > K) picks up at least 8 elements from H and at least 8 from K (see Exercise 35 in Chapter 24). Thus, |N(H > K)| $ 16 and is divisible by 8. So, |N(H > K)| 5 16, 56, or 112. Since the latter two cases yield normal subgroups, we may assume |N(H > K)| 5 16. If N(H > K) 5 H, then |H > K| 5 8, since N(H > K) contains at least 8 elements from K. So, we may assume that N(H > K) 2 H. Then, we may take A 5 N(H > K) and B 5 H. Use the Index Theorem. n5 5 6 and n3 5 10 or 40. If there are two Sylow 2-subgroups L2 and L29 whose intersection has order 4, show that N(L2 > L29) has index at most 5. Now use the Embedding Theorem. If n35 40, the union of all the Sylow subgroups has more than 120 elements. If n35 10, use the N/C Theorem to show that there is an element of order 6 and then use the Embedding Theorem and observe that A6 has no element of order 6. Let a be as in the proof of the Generalized Cayley Theorem. Then Ker a # H and |G/Ker a| divides |G:H|!. Now show |Ker a| 5 |H|. A subgroup of index 2 is normal. If H is a proper normal subgroup of S5, then H > A5 5 A5 or {e}. But H > A5 5 A5 implies H 5 A5, whereas H > A5 5 {e} implies H 5 {e} or |H| 5 2. (See Exercise 19 in Chapter 5.) Now use Exercise 70 in Chapter 9 and Exercise 48 in Chapter 5. By direct computation, show that PSL(2, Z7) has more than four Sylow 3-subgroups, more than one Sylow 7-subgroup, and more than one Sylow 2-subgroup. Hint: 1 4 d has order 3. Now use conjugation to find four other subgroups of order 3; Observe that c 1 5 5 5 d ` 5 7 and use conjugation to find another subgroup of order 7; observe that ` c 1 4 5 1 d ` 5 4 and use conjugation to find six more elements of order 4 (which guarobserve that ` c 3 5 antees that more than one Sylow 2-subgroup exists). Now argue as we did to show that A5 is simple. In the cases that the supposed normal subgroup N has order 2 or 4, show that in G/N, the Sylow 7-subgroup is normal. But then, G has a normal subgroup of order 14 or 28, which were already ruled out. Mimic Exercise 24. Suppose there is a simple group of order 60 that is not isomorphic to A5. The Index Theorem implies n2 2 1 or 3, and the Embedding Theorem implies n2 2 5. Thus, n2 5 15. Counting shows that there must be two Sylow 2-subgroups whose intersection has order 2. Now mimic the argument used in showing that there is no simple group of order 144 to show that the normalizer of this intersection has index 5, 3, or 1, but the Embedding Theorem and the Index Theorem rule these out. Suppose there is such a simple group G. Since the number of Sylow q-subgroups is 1 modulo q 2 2 2 and divides p , it must be p . Thus there are p (q 2 1) elements of order q in G. These elements, 2 together with the p elements in one Sylow p-subgroup, account for all p2q elements in G. Thus there cannot be another Sylow p-subgroup. But then the Sylow p-subgroup is normal in G. Consider the right regular representation of G. Let g be a generator of the Sylow 2-subgroup and suppose that |G| 5 2kn where n is odd. Then by Exercise 46 in Chapter 6 every cycle of the permutation Tg in the right regular representation of G has length 2k. This means that there are exactly n such cycles. Since each cycle is odd and there is an odd number of them, Tg is odd. This means that the set of even permutations in the regular representations has index 2 and is therefore normal. (See Exercise 19 in Chapter 5 and Exercise 7 in Chapter 9).

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Chapter 26 If you make a mistake, make amends. LOU HOLTZ

1. u , u because u is obtained from itself by no insertions; if v can be obtained from u by inserting or deleting words of the form xx21 or x21x, then u can be obtained from v by reversing the procedure; if u can be obtained from v and v can be obtained from w, then u can be obtained from w by obtaining first v from w and then u from v. 3. b(a2N) 5 b(aN)a 5 a3bNa 5 a3b(aN) 5 a3a3bN 5 a6bN 5 a6Nb 5 a2Nb 5 a2bN b(a3N) 5 b(a2N)a 5 a2bNa 5 a2b(aN) 5 a2a3bN 5 a5bN 5 a5Nb 5 aNb 5 abN b(bN) 5 b2N 5 N b(abN) 5 baNb 5 a3bNb 5 a3b2N 5 a3N b(a2bN) 5 ba2Nb 5 a2bNb 5 a2b2N 5 a2N b(a3bN) 5 ba3Nb 5 abNb 5 ab2N 5 aN 5. Let F be the free group on {a1, a2, . . . , an}. Let N be the smallest normal group containing {w1, w2, . . . , wt} and let M be the smallest normal subgroup containing {w1, w2, . . . , wt, wt11, . . . , wt1k}. Then F/N < G and F/M < G. The homomorphism from F/N to F/M given by aN → aM induces a homomorphism from G onto G. To prove the corollary, observe that the theorem shows that K is a homomorphic image of G, so |K| # |G|. 7. Clearly, a and ab belong to a, b, so a, ab # a, b. Now show that a and b belong to a, ab. 9. Show that |G| # 2n and that Dn satisfies the relations that define G. 11. Since x2 5 y2 5 e, we have (xy)21 5 y21x21 5 yx. Also, xy 5 z21yz, so (xy)21 5 (z21yz)21 5 z21y21z 5 z21yz 5 xy. 13. a. b6 b. b7a 15. Center is x2. |xy| 5 8. 17. Use the fact that the mapping from G onto G/N given by x → xN is a homomorphism. 19. For H to be a normal subgroup we must have yxy21 [ H 5 {e, y3, y6, y9, x, xy3, xy6, xy9}. But yxy21 5 yxy11 5 (yxy)y10 5 xy10. 21. 6; the given relations imply that a2 5 e. G is isomorphic to Z6. 23. 1, 2, and ` 25. ab 5 c 1 abc21 5 e cd 5 a 1 (abc21)cd 5 ae 1 bd 5 e 1 d 5 b21 da 5 b 1 bda 5 b2 1 ea 5 b2 1 a 5 b2 ab 5 c 1 b3 5 c So G 5 b. bc 5 d 1 bb3 5 b21 1 b5 5 e. So |G| 5 1 or 5. But Z5 satisfies the defining relations with a 5 1, b 5 3, c 5 4, and d 5 2. 27. Z6

Chapter 27 If at first you don’t succeed—that makes you about average. BRADENTON, Florida Herald

1. If T is a distance-preserving function and the distance between points a and b is positive, then the distance between T(a) and T(b) is positive. 3. See Figure 1.5.

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A35

5. 12 7. 4n 9. a. Z2 b. Z2 % Z2 c. G % Z2, where G is the plane symmetry group of a circle. 11. 6 13. An inversion in R3 leaves only a single point fixed, whereas a rotation leaves a line fixed. 15. In R4, a plane is fixed. In Rn, a hyperplane of dimension n 2 2 is fixed. 17. Create a coordinate system for the plane. Let T be an isometry; p, q, and r the three noncollinear points; and s any other point in the plane. Then the quadrilateral determined by T( p), T(q), T(r), and T(s) is congruent to the one formed by p, q, r, and s. Thus, T(s) is uniquely determined by T ( p), T(q), and T (r). 19. a rotation

Chapter 28 The thing that counts is not what we know but the ability to use what we know. LEO L. SPEARS

1. 3. 5. 7. 9. 11. 13. 15. 17.

19.

Try xnym → (n, m). xy Use Figure 28.9. x2yzxz 5 x2yx21 5 x2x21y 5 xy x23zxyz 5 x23x21y 5 x24y A subgroup of index 2 is normal. a. V b. I c. II d. VI e. VII f. III cmm a. p4m b. p3 c. p31m d. p6m The principal purpose of tire tread design is to carry water away from the tire. Patterns I and III do not have horizontal reflective symmetry. Thus these designs would not carry water away equally on both halves of the tire. a. VI b. V c. I d. III e. IV f. VII g. IV

Chapter 29 With every mistake we must surely be learning. GEORGE HARRISON, “While My Guitar Gently Weeps,” The White Album

1. 3. 5. 7. 9.

6 30 13 45 126 1 11. (n6 1 2 ? n 1 2 ? n2 1 n3 ) 6 13. For the first part, see Exercise 11 in Chapter 6. For the second part, try D4. 15. The kernel is {R0, R180, H, V}.

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Chapter 30 I am not bound to please thee with my answers. SHAKESPEARE,

1. 3. 5. 7. 11.

13. 15.

17. 19.

21. 23. 25. 27. 29. 31. 33. 35.

The Merchant of Venice

4 * (b, a) (m/2) * {3 * [(a, 0), (b, 0)], (a, 0), (e, 1), 3 * (a, 0), (b, 0), 3 * (a, 0), (e, 1)} a3b Both yield paths from e to a3b. Say we start at x. Then we know the vertices x, xs1, xs1s2, . . . , xs1s2 . . . sn21 are distinct and x 5 xs1s2 . . . sn. So if we apply the same sequence beginning at y, then cancellation shows that y, ys1, ys1s2, . . . , ys1s2 . . . sn21 are distinct and y 5 ys1s2 . . . sn. If there were a Hamiltonian path from (0, 0) to (2, 0), there would be a Hamiltonian circuit in the digraph, since (2, 0) 1 (1, 0) 5 (0, 0). a. If s1, s2, . . . , sn21 traces a Hamiltonian path and sisi11 . . . sj 5 e, then the vertex s1s2 . . . si21 appears twice. Conversely, if sisi11 . . . sj 2 e, then the sequence e, s1, s1s2, . . . , s1s2 . . . sn21 yields the n vertices (otherwise, cancellation gives a contradiction). b. This follows directly from part a. The sequence traces the digraph in a clockwise fashion. Abbreviate (a, 0), (b, 0), and (e, 1) by a, b, and 1, respectively. A circuit is 4 * (4 * 1, a), 3 * a, b, 7 * a, 1, b, 3 * a, b, 6 * a, 1, a, b, 3 * a, b, 5 * a, 1, a, a, b, 3 * a, b, 4 * a, 1, 3 * a, b, 3 * a, b, 3 * a, b. Abbreviate (R90, 0), (H, 0), and (R0, 1) by R, H, and 1, respectively. A circuit is 3 * (R, 1, 1), H, 2 * (1, R, R), R, 1, R, R, 1, H, 1, 1. Abbreviate (a, 0), (b, 0), and (e, 1) by a, b, and 1, respectively. A circuit is 2 * (1, 1, a), a, b, 3 * a, 1, b, b, a, b, b, 1, 3 * a, b, a, a. Abbreviate (r, 0), ( f, 0), and (e, 1) by r, f, and 1, respectively. Then the sequence is r, r, f, r, r, 1, f, r, r, f, r, 1, r, f, r, r, f, 1, r, r, f, r, r, 1, f, r, r, f, r, 1, r, f, r, r, f, 1. m * ((n 2 1) * (0, 1), (1, 1)) Abbreviate (r, 0), ( f, 0), and (e, 1) by r, f, and 1, respectively. A circuit is 1, r, 1, 1, f, r, 1, r, 1, r, f, 1. 5* [3 * (1, 0), (0, 1)], (1, 0) 12 * ((1, 0), (0, 1)). In the proof of Theorem 30.3, we used the hypothesis that G is Abelian in two places: We needed H to satisfy the induction hypothesis, and we needed to form the factor group G/H. Now, if we assume only that G is Hamiltonian, then H also is Hamiltonian and G/H exists.

Chapter 31 We must view with profound respect the infinite capacity of the human mind to resist the introduction of useful knowledge. THOMAS R. LOUNSBURY

1. 3. 5. 7. 9.

wt(0001011) 5 3; wt(0010111) 5 4; wt(0100101) 5 3; etc. 1000110; 1110100 000000, 100011, 010101, 001110, 110110, 101101, 011011, 111000 Not all single errors can be detected. Observe that a vector has even weight if and only if it can be written as a sum of an even number of vectors of weight 1. 11. No, by Theorem 31.3.

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13. 0000000, 1000111, 0100101, 0010110, 0001011, 1100010, 1010001, 1001100, 0110011, 0101110, 0011101, 1110100, 1101001, 1011010, 0111000, 1111111; 1 1 1 H 5 P0 1 0 0

1 0 1 1 0 1 0

1 1 0 1p ; 0 0 1

yes. 15. Suppose that u is decoded as v and that x is the coset leader of the row containing u. Coset decoding means v is at the head of the column containing u. So, x 1 v 5 u and x 5 u 2 v. Now suppose u 2 v is a coset leader and u is decoded as y. Then y is at the head of the column containing u. Since v is a code word, u 5 u 2 v 1 v is in the row containing u 2 v. Thus u 2 v 1 y 5 u and y 5 v. 17. 000000, 100110, 010011, 001101, 110101, 101011, 011110, 111000 1 0 1 H5O 1 0 0

19.

21. 23. 25. 27. 29.

1 1 0 0 1 0

0 1 1 o 0 0 1

001001 is decoded as 001101 by all four methods. 011000 is decoded as 111000 by all four methods. 000110 is decoded as 100110 by all four methods. Since there are no code words whose distance from 100001 is 1 and three whose distance is 2, the nearest-neighbor method will not decode or will arbitrarily choose a code word; parity-check matrix decoding does not decode 100001; the standard-array and syndrome methods decode 100001 as 000000, 110101, or 101011, depending on which of 100001, 010100, or 001010 is a coset leader. For any received word w, there are only eight possibilities for wH. But each of these eight possibilities satisfies condition 2 or the first portion of condition 39 of the decoding procedure, so decoding assumes that no error was made or one error was made. There are 34 code words and 36 possible received words. No; row 3 is twice row 1. No. For if so, nonzero code words would be all words with weight at least 5. But this set is not closed under addition. Use Exercise 24, together with the fact that the set of code words is closed under addition. Abbreviate the coset a 1 x2 1 x 1 1 with a. The following generating matrix will produce the desired code: c

1 0 1 1 x d. 0 1 x x11 x11

31. Use Exercise 14. 33. Let c, c9 [ C. Then, c 1 (v 1 c9) 5 v 1 c 1 c9 [ v 1 C and (v 1 c) 1 (v 1 c9) 5 c 1 c9 [ C, so the set C < (v 1 C) is closed under addition.

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35. If the ith component of both u and v is 0, then so is the ith component of u 2 v and au, where a is a scalar.

Chapter 32 Wisdom rises upon the ruins of folly. THOMAS FULLER,

Gnomologia

1. Note that f(1) 5 1. Thus f(n) 5 n. Also, 1 5 f(1) 5 f(nn21) 5 f(n)f(n21) 5 nf(n21), so that 1/n 5 f(n21). 3. If a and b are automorphisms of E fixing F, so are a21 and ab. 5. If a and b are fixed by elements of H, so are a 1 b, a 2 b, a ? b, and a/b. 7. It suffices to show that each member of Gal(K/F) defines a permutation on the ai’s. Let a [ Gal(K/F) and write f(x) 5 cnxn 1 cn21xn21 1 . . . 1 c0 5 cn(x 2 a1)(x 2 a2) . . . (x 2 an).

9.

11. 13. 15. 17. 19. 21. 23. 25. 27.

Then f(x) 5 a(f(x)) 5 cn(x 2 a(a1))(x 2 a(a2)) ? ? ? (x 2 a(an)). Thus, f(ai) 5 0 implies ai 5 a(aj) for some j, so that a permutes the ai’s. f6(v) 5 v729 5 v. f3(v 1 v21) 5 v27 1 v227 5 v21 1 v. f2(v3 1 v5 1 v6) 5 v27 1 v45 1 v54 5 v6 1 v3 1 v5. Recall that A4 has no subgroup of order 6. (See Example 13 in Chapter 9.) Use Sylow’s Theorem. 3 3 3 Let v be a primitive cube root of 1. Then Q , Q(" 2 ) , Q(v," 2 ) and Q(" 2 ) is not the splitting field of a polynomial in Q[x]. Use the lattice of Z10. Z6 (Be sure you know why the group is cyclic.) See Exercise 21 in Chapter 25. Use Exercise 33 in Chapter 24. Use Exercise 40 in Chapter 10. Since K/N v G/N, for any x [ G and k [ K, there is a k9 [ K such that k9N 5 (xN)(kN) (xN)21 5 xNkNx21N 5 xkx21N. So, xkx21 5 k9n for some n [ N. And since N # K, we have k9n [ K.

Chapter 33 All wish to posses knowledge, but few, comparatively speaking, are willing to pay the price. JUVENAL

1. x 2 2 x 1 1. 3. Over Z, x8 2 1 5 (x 2 1)(x 1 1)(x2 1 1) (x4 1 1). Over Z2, x2 1 1 5 (x 1 1)2 and x4 1 1 5 (x 1 1)4. So, over Z2, x8 2 1 5 (x 1 1)8. Over Z3, x2 1 1 is irreducible, but x4 1 1 factors into irreducibles as (x2 1 x 1 2) (x2 2 x 2 1). So, x8 2 1 5 (x 2 1)(x 1 1) (x2 1 1)(x2 1 x 1 2)(x2 2 x 2 1). Over Z5, x2 1 1 5 (x 2 2)(x 1 2), x4 1 1 5 (x2 1 2)(x2 2 2), and these last two factors are irreducible. So, x8 2 1 5 (x 2 1)(x 1 1) (x 2 2)(x 1 2)(x2 1 2)(x2 2 2). 5. Let v be a primitive nth root of unity. We must prove vv2 . . . vn 5 (21)n11. Observe that vv2 . . . vn 5 vn(n11)/2. When n is odd, vn(n11)/2 5 (vn)(n11)/2 5 1(n11)/2 5 1. When n is even, (vn/2)n11 5 (21)n11 5 21. 7. If [F : Q] 5 n and F has infinitely many roots of unity, then there is no finite bound on their multiplicative orders. Let v be a primitive mth root of unity in F such that f(m) . n. Then [Q(v) : Q] 5 f(m). But F $ Q(v) $ Q implies [Q(v) : Q] # n.

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A39

9. Let 2n 1 1 5 q. Then 2 [ U(q) and 2n 5 q 2 1 5 21 in U(q) implies that |2| 5 2n. So, by Lagrange’s Theorem, 2n divides |U(q)| 5 q 2 1 5 2n. 11. Let v be a primitive nth root of unity. Then 2nth roots of unity are 61, 6v, . . . , 6vn21. These are distinct, since 21 5 (2vi)n, whereas 1 5 (v i)n. 13. First observe that deg F2n(x) 5 f(2n) 5 f(n) and deg Fn(2x) 5 deg Fn(x) 5 f(n). Thus, it suffices to show that every zero of Fn(2x) is a zero of F2n(x). But the fact that v is a zero of Fn(2x) means that |2v| 5 n, which in turn implies that |v| 5 2n. 2p 15. Let G 5 Gal(Q(v)/Q) and H1 be the subgroup of G of order 2 that fixes cos( n ). Then, by . . . induction, G/H1 has a series of subgroups H1/H1 , H2/H1 , , Ht /H1 5 G/H1, so that |Hi11/H1 : Hi/H1| 5 2. Now observe that |Hi11/H1 : Hi /H1| 5 |Hi11/Hi|. 17. Instead, prove that Fn(x)Fpn(x) 5 Fn(xp). Since both sides are monic and have degree pf(n), it suffices to show that every zero of Fn(x)Fpn(x) is a zero of Fn(xp). If v is a zero of Fn(x), then |v| 5 n. By Theorem 4.2, |vp| 5 n also. Thus, v is a zero of Fn(xp). If v is a zero of Fpn(x), then |v| 5 pn and therefore |v p| 5 n. 19. Use Theorem 33.4 and Theorem 32.1. 21. v → v4, v → v21, v → v24

Supplementary Exercises for Chapters 24–33 For those who keep trying, failure is temporary. FRANK TYGER

1. Z6 3. Let |G| 5 315 and let H be a Sylow 3-subgroup and K a Sylow 5-subgroup. If H v G, then HK 5 45. If H is not normal, then by Sylow’s Third Theorem, |G/N(H)| 5 7, so that |N(H)| 5 45. 5. Observe that K # N(H) implies that HK is a group of order 245. Now, use Sylow’s Third Theorem. 7. Note that gKg21 # gHg21 5 H. Now use the corollary to Sylow’s Third Theorem. 9. Use the same proof as for Exercise 55 in Chapter 9. 11. Since n7 5 8, we know by the Embedding Theorem (Chapter 25) that G # A8. But A8 does not have an element of order 21. 13. Let G be a non-Abelian group of order 105. By Theorem 9.3, G/Z(G) is not cyclic. So |Z(G)| 2 3, 7, 15, 21, or 35. This leaves only 1 or 5 for |Z(G)|. Let H, K, and L be Sylow 3-, Sylow 5-, and Sylow 7-subgroups of G, respectively. Now, counting shows that K v G or L v G. Thus, |KL| 5 35 and KL is a cyclic subgroup of G. But, KL has 24 elements of order 35 (since |U(Z35)| 5 24). Thus, a counting argument shows that K v G and L v G. Now, |HK| 5 15 and HK is a cyclic subgroup of G. Thus, HK # C(K) and KL # C(K). This means that 105 divides |C(K)|. So K # Z(G). 15.

17. It suffices to show that x travels by a implies xab21 travels by a (for we may successively replace x by xab21). If xab21 traveled by b, then the vertex xa would appear twice in the circuit. 19. a. {00, 11} b. {000, 111} c. {0000, 1100, 1010, 1001, 0101, 0110, 0011, 1111} d. {0000, 1100, 0011, 1111} 21. The mapping Tv: Fn → {0, 1} given by Tv(u) 5 u ? v is an onto homomorphism. So |Fn/Ker Tv| 5 2. 23. It follows from Exercise 18 that if C is an (n, k) linear code, then C> is an (n, n 2 k) linear code. Thus, in this problem, k 5 n 2 k. To prove the second claim, use Exercise 18, Exercise 21, the definition of C>, and the hypothesis that C> 5 C.

Text Credits Page xi: Copyright © 1966 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission. Page 14: “Brain Boggler” by Maxwell Carver, Copyright © 1988 by Discover Magazine. Used by permission. Page 52: GRE materials selected from the GRE Practice Exam, Question 9 by Educational Testing Service. Reprinted by permission of Educational Testing Service, the copyright owner. Page 70: “A Programmer’s Lament,” from The Compleat Computer, Second Edition, 1983, by Dennie L. VanTassel and Cynthia VanTassel. Reprinted by permission of Dennie L. VanTassel. Page 113: Copyright © Piet Hein Grooks. Reprinted with kind permission from Piet Hein a/s, DK-5500 Middelfart, Denmark. Page 173: From Gina Kolata, “Hitting the High Spots of Computer Theory,” New York Times, December 13, 1994. Reprinted with permission of PARS International. Page 226: Copyright © 1969 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission. Page 269: Copyright © Piet Hein Grooks. Reprinted with kind permission from Piet Hein a/s, DK-5500 Middelfart, Denmark. Page 287: Copyright © 1965 (Renewed) Stony/ATV Tunes LLC. All rights administered by Sony/ATV Music Publishing, 8 Music Square West, Nashville, TN 37203. All rights reserved. Used by permission. Page 339: Quote from Math’s Hidden Woman, Nova Online, http://www.pbs .org/wgbh/nova/proof/germain.html (accessed Nov 5, 2008). Page 459: Copyright © 1994. Reprinted with permission of the author and the publisher from PARADE, December 11, 1994. Page 467: Adaptation of figure from Dorothy K. Washburn and Donald W. Crowe. Symmetries of Culture: Theory and Practice of Plane Pattern Analysis. Copyright © 1988 by the University of Washington Press. Used by permission. Page 535–536: Adapted with permission from “The Ubiquitous Reed-Solomon Codes” by Barry A. Cipra, Siam News. January 1993. Page 572: Copyright © Piet Hein Grooks. Reprinted with kind permission from Piet Hein a/s, DK-5500 Middelfart, Denmark.

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Index of Mathematicians

(Biographies appear on pages in boldface.)

Abbati, Pietro, 141 Abel, Niels, 32, 39, 325, 419, 552 Adleman, L., 162, 164, 165, 173–174 Allenby, R., 123 Artin, Emil, 321, 352, 353 Artin, Michael, 352 Aschbacher, Michael, 424, 431, 434, 452 Bell, E.T., 383 Berlekamp, Elwyn, 536 Bhargava, Manjul, 571 Bieberbach, L., 476 Birkhoff, Garrett, 304 Boole, George, 245 Brauer, Richard, 421, 431 Burnside, William, 421, 488, 497, 560

Eisenstein, Ferdinard, 309, 570 Erdös, Paul, 517 Escher, M.C., 484, 485, 509–511 Euclid, 6, 393 Euler, Leonhard, 39, 44, 154, 325, 339, 370, 517 Feit, Walter, 421, 424, 431, 436, 497, 553 Fermat, Pierre, 143, 325, 326 Fields, John, 430 Fischer, Bernd, 424 Fourier, Joseph, 199 Fraenkel, Abraham, 237 Frobenius, Georg, 174, 408, 488

Cauchy, Augustin-Louis, 98, 104, 121, 187, 199, 266, 325, 354, 408 Cayley, Arthur, 31, 89, 90, 95, 126, 137, 382, 444, 498 Chevalley, Claude, 431 Cole, Frank, 421, 430 Conway, John H., 424, 486 Courant, Richard, 353 Crowe, Donald, 467

Galois, Evariste, 40, 123, 138, 154, 178, 199, 369, 382, 383, 420, 421, 429, 552, 553, 566, 568, 570, 571 Gauss, Carl, 39, 112, 121, 160, 274, 307, 309, 313, 325, 334, 339, 378, 383, 394, 561, 564, 567, 568, 570, 571 Germain, Sophie, 339 Gersonides, 282 Gorenstein, Daniel, 423, 435, 534 Griess, Robert, 424

Davenport, Harold, 452 da Vinci, Leonardo, 455 Dedekind, Richard, 237, 274, 383, 570 De Morgan, Augustus, 12, 90 de Seguier, J. A., 62 Dickson, L.E., 392, 421, 431 Dirichlet, Peter, 325, 570 Dyck, Walther, 40, 442

Hall, Marshall, 434, 452 Hall, Philip, 452, 560 Hamilton, William Rowen, 196, 502, 516 Hamming, Richard, 520, 521, 535, 542 Hardy, G.H., 452 Hermite, Charles, 370 Herstein, I.N., 248, 435

A43

A44

INDEX OF MATHEMATICIANS

Hilbert, David, 237, 275, 325, 352, 353, 475 Hölder, Otto, 131, 180, 217, 420, 430 Holst, Elling, 419

Pascal, Blaise, 12 Pless, Vera, 544 Poincaré, Henri, 369, 451 Pólya, George, 484, 485

Jacobson, Nathan, 261 Jordan, Camille, 180, 200, 207, 217, 383, 419, 420, 421, 429

Rankin, R.A., 509 Reed, Irving, 535, 536 Riemann, Bernhard, 570 Rivest, R., 162, 164, 173 Ruffini, Paolo, 102, 142

Kaplansky, Irving, 381 Klein, Felix, 275, 429, 430 Kline, Morris, 274 Knuth, Donald, 434, 452 Kronecker, Leopold, 218, 325, 354, 369 Kummer, Ernst, 325, 326, 369, 570 Lagrange, Joseph-Louis, 39, 121, 141, 154, 199, 339 Lamé, Gabriel, 325, 326 Landau, Edmund, 326 Lange, Serge, 321 Laplace, Pierre-Simon, 121, 154 Larson, Loren, 108 Legendre, Adrien-Marie, 199, 325 Lie, Sophus, 419 Lindemann, Ferdinand, 360, 369, 370, 394 Liouville, Joseph, 370 Mac Lane, Saunders, 304 MacWilliams, Jessie, 543 Mathieu, Emile, 421 Maurolycus, Francisco, 12 McElicee, Robert, 452, 535, 536 Miller, G.A., 138, 174, 421 Miyaoka, Yoichi, 326 Moore, E.H., 131, 383 Motzkin, T., 333 Netto, Eugen, 89 Newton, Isaac, 39, 154, 339 Noether, Emmy, 275, 330, 556

Schattschneider, Doris, 467 Shamir, Adi, 162, 164, 165, 173 Shannon, Claude, 535 Singer, Richard, 331 Slepian, David, 531 Sloane, Neil, 543 Smith, Stephen, 424 Solomon, Gustave, 535, 536 Steinitz, Ernst, 285, 375 Sylow, Ludwig, 91, 406, 419 Sylvester, J.J., 89–90, 137 Taylor, Richard, 326, 340 Taussky-Todd, Olga, 353 Thompson, John G., 421, 423, 424, 431, 435, 436, 452, 453, 497, 556, 560 van der Waerden, B.L., 217 Verhoeff, J., 110, 111 Wantzel, Pierre, 394 Weber, Heinrich, 40, 44, 251 Weyl, Hermann, 275, 455 Wiles, Andrew, 326, 340, 430, 571 Zariski, Oscar, 435 Zelmanov, Efim, 497 Zierler, N., 534 Zorn, Max, 248

Index of Terms

Abel Prize, 39, 436 Abelian group, 32, 41 Addition modulo n, 7 Additive group of integers modulo n, 42 Algebraic closure, 377, 378 element, 370 extension, 370 Algebraically closed field, 378 Alternating group, 106 Annihilator, 272 Arc, 498 Ascending chain condition, 329, 336 Associates, 322 Associativity, 32, 41 Automorphism(s) Frobenius, 389 group, 131, 509 group of E over F, 546 inner, 130 of a group, 130 Axioms for a group, 41 for a ring, 237 for a vector space, 345 Basis for a vector space, 347 Binary code, 523 operation, 40 strings, 161 Boolean ring, 245 Burnside’s Theorem, 489 Cancellation property for groups, 48 property for integral domains, 250 Cauchy’s Theorem, 187, 408 Cayley digraph, 498 Cayley table, 31 Cayley’s Theorem, 126 generalized, 426

Center of a group, 62 of a ring, 243 Centralizer of an element, 64 of a subgroup, 66 Characteristic of a ring, 252 Characteristic subgroup, 174 Check digit, 7 Check-digit scheme, 110 Chinese Remainder Theorem for Rings, 341 Circle in F, 394 Class equation, 404 Closure, 31, 40 Code binary, 523 dual of, 573 Hamming, 520 (n,k) linear, 523 self-dual, 573 systematic, 526 ternary, 524 word, 520, 523 Cole Prize, 321, 352, 392, 424, 430, 434, 436, 571 Commutative diagram, 208 Commutative operation, 32 Commutator subgroup, 174 Composition of functions, 19 Composition factors, 420 Conjugacy class, 91, 403 Conjugate elements, 403 subgroups, 91, 408 Conjugation, 126 Constant polynomial, 295 Constructible number, 394 Constructible regular n-gons, 566 Content of a polynomial, 306 Coset decoding, 531

A45

A46

INDEX OF TERMS

leader, 531 left, 138 representative, 138 right, 138 Crystallographic groups, 467 Crystallographic restriction, 473 Cube, rotation group of, 147 Cycle m-, 98 notation, 98 Cyclic group, 72 rotation group, 34 subgroup, 61 Cyclotomic extension, 562 polynomial, 310, 562 Decoding coset, 531 maximum-likelihood, 518 nearest neighbor, 520 parity-check matrix, 528 Degree of a over F, 372 of an extension, 372 of a polynomial, 295 rule, 301 DeMoivre’s Theorem, 13 Derivative, 362 Determinant, 43 Diagonal of G % G, 168 Digital signatures, 165 Dihedral groups, 31, 32 Dimension of a vector space, 349 Direct product of groups external, 155 internal, 188, 190 Direct sum of groups, 192 of rings, 239 Dirichlet’s Theorem, 228 Discrete frieze group, 461 Distance between vectors, 524 Divides, 238, 298 Division algorithm for F[x], 296 for Z, 3 Divisor, 3

Domain Euclidean, 331 integral, 249 Noetherian, 330 unique factorization, 328 Doubling the cube, 393, 395 Dual code, 573 Eisenstein’s criterion, 309 Element(s) algebraic, 370 conjugate, 403 degree of, 372 fixed by f, 489 idempotent, 255 identity, 31, 41, 238 inverse, 31, 41 nilpotent, 255 order of, 57 primitive, 376 square, 195 transcendental, 370 Embedding Theorem, 427 Empty word, 438 Equivalence class, 16 Equivalence relation, 16 Equivalent under group action, 487 Euclidean domain, 331 Euclid’s Lemma, 6 generalization of, 23 Euler phi-function, 79 Even permutation, 105 Exponent of a group, 175 Extension algebraic, 370 cyclotomic, 562 degree, 372 field, 354 finite, 372 infinite, 372 simple, 370 transcendental, 370 External direct product, 155 Factor group, 180 of a ring element, 238 ring, 263 Factor Theorem, 298

INDEX OF TERMS

Feit-Thompson Theorem, 421, 423, 436, 497, 553 Fermat prime, 568 Fermat’s Last Theorem, 325–327 Fermat’s Little Theorem, 143 Field algebraic closure of, 377, 378 algebrically closed, 378 definition of, 250 extension, 354 fixed, 546 Galois, 383 of quotients, 285 perfect, 364 splitting, 356 Fields Medal, 423, 430, 436, 497 Finite dimensional vector space, 349 Finite extension, 372 First Isomorphism Theorem for groups, 207 for rings, 283 Fixed field, 546 Free group, 439 Frieze pattern, 461 Frobenius map, 289, 389 Function composition, 19 definition of, 18 domain, 18 image under, 18 one-to-one, 19 onto, 20 range, 18 Fundamental region, 473 Fundamental Theorem of Algebra, 378 of Arithmetic, 6 of Cyclic Groups, 77 of Field Theory, 354 of Finite Abelian Groups, 218 of Galois Theory, 550 of Group Homomorphisms, 207 of Ring Homomorphisms, 284 GAP, 109 G/Z Theorem, 186 Galois field, 383 group, 546, 558 Gaussian integers, 241, 332

A47

Gauss’s Lemma, 307 Generating region of a pattern, 473 Generator(s) of a cyclic group, 61, 72 of a group, 47 in a presentation, 441 Geometric constructions, 393 Glide-reflection, 454 nontrivial, 464 trivial, 464 Greatest common divisor, 5 Group Abelian, 32, 41 action, 493 alternating, 106 automorphism, 131, 509 automorphism of, 130 center of, 62 color graph of a , 499 commutative, 32 composition factors, 420 crystallographic, 467 cyclic 34, 61, 72 definition, 41 dicyclic, 445, 450 dihedral, 31, 32 discrete frieze group, 461 factor, 180 finite, 57 free, 439 frieze, 461 Galois, 546, 558 general linear, 43 generator(s), 47, 61, 72, 441 Hamiltonian, 514 Heisenberg, 54 homomorphism of, 200 icosahedral, 430, 457 infinite dihedral, 446 inner automorphism, 131 integers mod n, 42 isomorphic, 123 isomorphism, 123 non-Abelian, 32, 41 octahedral, 457 order of, 57 p- 404, 417 permutation, 95 presentation, 441 quarternions, 91, 196, 442

A48

INDEX OF TERMS

quotient, 180 representation, 211 simple, 420 solvable, 553 space, 475 special linear, 45 symmetric, 97 symmetry, 33, 34, 453 tetrahedral, 457 of units, 243 wallpaper, 467 Half-turn, 463 Hamiltonian circuit, 503 group, 514 path, 503 Hamming code, 520 distance, 524 weight of a code, 524 weight of a vector, 524 Homomorphism(s) Fundamental Theorem of, 207, 284 kernel of, 200 of a group, 200 natural, 210, 284 of a ring, 280 Ideal annihilator, 272 definition of, 262 finitely generated, 336 generated by, 263 maximal, 267 nil radical of, 272 prime, 267 principal, 263 product of, 270 proper, 262 sum of, 270 test, 262 trivial, 263 Idempotent, 255 Identity element, 31, 41, 238 Index of a subgroup, 142 Index Theorem, 426 Induction first principle of, 13 second principle of, 14 Inner automorphism, 130

Integral domain, 249 Internal direct product, 188, 190 International standard book number, 24 Inverse element, 31, 41 Inverse image, 204 Inversion, 135 Irreducibility tests, 306, 308 Irreducible element, 322 Irreducible polynomial, 305 ISBN, 24 Isometry, 453 Isomorphism(s) class, 218 First Theorem for groups, 207 First Theorem for rings, 283 of groups, 123 of rings, 280 Second Theorem for groups, 214 Second Theorem for rings, 341 Third Theorem for groups, 214 Third Theorem for rings, 341 Kernel of a homomorphism, 200 of a linear transformation, 351 Key, 162 Kronecker’s Theorem, 354 Lagrange’s Theorem 141 Latin square, 53 Lattice diagram, 80 of points, 473 unit, 473 Leading coefficient, 295 Least common multiple, 6 Left regular representation, 127 Line in F, 394 Linear code, 523 combination, 347 transformation, 351 Linearly dependent vectors, 347 Linearly independent vectors, 347 Mathematical induction First Principle, 13 Second Principle, 14 Mapping, 18

INDEX OF TERMS

Matrix addition, 42 determinant of, 43 multiplication, 43 standard generator, 526 Maximal, ideal, 267 subgroup, 232 Maximum-likelihood decoding, 518 Measure, 331 Minimal polynomial, 371 Mirror, 454 Mod p Irreducibility Test, 308 Modular arithmetic, 7 Monic polynomial, 295 Monster, 424, 556 Multiple, 3 Multiple zeros, 363 Multiplication modulo n, 7 Multiplicity of a zero, 298 Natural homomorphism, 210, 281, 284 Natural mapping, 208 N/C Theorem, 209 Nearest-neighbor decoding, 520 Nilpotent element, 255 Nil radical, 272 Noetherian domain, 330 Norm, 323 Normal subgroup, 178 Normal Subgroup Test, 179 Normalizer, 91 Odd permutation, 105 Operation associative, 41 binary, 40 commutative, 32 preserving mapping, 123 table, 31 Opposite isometry, 454 Orbit of a point, 145 Orbit-Stabilizer Theorem, 146 Order or a group, 57 of an element, 57 Orthogonality Relation, 530 PID, 299 Parity-check matrix, 528

Partition of a set, 17 of an integer, 219 Perfect field, 364 Permutation definition of, 95 even, 105 group, 95 odd, 105 p-group 404, 417 Phi-function, Euler, 79 Plane of F, 394 Plane symmetry, 33 Polynomial(s) alternating, 106 constant, 295 content of, 306 cyclotomic, 310, 562 degree of, 295 derivative of, 362 Galois group of, 558 irreducible, 305 leading coefficient of, 295 minimal, 371 monic, 295 primitive, 306 reducible, 305 relatively prime, 303 ring of, 293 splits, 356 symmetric, 106 zero of, 298 Prime element of a domain, 322 ideal, 267 integer, 3 relatively, 5, 303 subfield, 285 Primitive element, 376 Element Theorem, 375 nth root of unity 299, 562 polynomial, 306 Principal ideal domain 271, 299 Principal ideal ring, 290 Projection, 212 Proper ideal, 262 Proper subgroup, 58 Pullback, 204

A49

A50

INDEX OF TERMS

Quaternions, 91, 196, 442 Quotient, 4, 297 Quotient group, 180 Quotients, field of, 285 Range, 18 Rational Root Theorem, 318 Reducible polynomial, 305 Reflection, 34, 454 Relation equivalence, 16 in a presentation, 441 Relatively prime, 5, 303 Remainder, 4, 297 Remainder Theorem, 298 Ring(s) Boolean, 245 center of, 243 characteristic of, 252 commutative, 238 definition of, 237 direct sum of, 239 factor, 263 homomorphism of, 280 isomorphism of, 280 of polynomials, 293 with unity, 238 RSA public encryption, 164 Rubik’s cube, 110 Scalar, 345 Scalar multiplication, 345 Self dual code, 573 Sicherman dice, 315 Simple extension, 370 Simple group, 420 Socks-Shoes Property, 50 Solvable by radicals, 552 Solvable group, 553 Spanning set, 347 Splitting field, 356 Squaring the circle, 393, 396 Stabilizer of a point, 115, 145 Standard array, 531 Standard generator matrix, 526 Subcode, 537 Subfield Test, 256 Subgroup(s) centralizer, 66 characteristic, 174

commutator, 174 conjugate, 91, 408 cyclic, 61 definition of, 58 diagonal, 168 Finite Test, 61 generated by a, 61 index of, 142 lattice, 80 maximal, 232 nontrivial, 58 normal, 178 One-Step Test, 59 proper, 58 Sylow p- , 407 torsion, 92 trivial, 58 Two-Step Test, 60 Subring definition of, 240 Test, 240 Trivial, 241 Subspace, 346 Subspace spanned by vectors, 347 Subspace Test, 349 Sylow p-subgroup, 407 Sylow test for nonsimplicity, 425 Sylow Theorems, 406, 408, 409 Symmetric group, 97 Symmetries of a square, 29 Symmetry group, 33, 34, 453 Syndrome of a vector, 533 Systematic code, 526 Torsion subgroup, 92 Transcendental element, 370 Transcendental extension, 370 Translation, 45, 454 Transposition, 103 Trisecting an angle, 393, 396 UFD, 328 Unique factorization domain, 328 Unique factorization theorem for a PID, 329 for D[x], 334 for F[x], 331 for Z, 6 for Z[x], 313 in a Euclidean domain, 333

INDEX OF TERMS

Unity, 238 Universal Factor Group Property, 440 Universal Mapping Property, 439 Universal Product Code, 9 Vector, 345 Vector space basis of, 347 definition of, 345 dimension of, 349 finite dimensional, 349 infinite dimensional, 349 spanned by a set, 347 trivial, 349 Vertex of a graph, 498

Wallpaper groups, 467 Weight of a vector, 524 Weighting vector, 10 Weird dice, 315 Well-defined function, 201 Well-ordering principle, 3 Word code, 520, 523 empty, 438 in a group, 438 Zero multiple, 363 multiplicity of, 298 of a polynomial, 298 Zero-divisor, 249

A51

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Essential Theorems in Abstract Algebra Theorem 3.1 One-Step Subgroup Test A nonempty subset H of a group G is a subgroup of G if ab1 is in H whenever a and b are in H. Theorem 4.3 Fundamental Theorem of Cyclic Groups Every subgroup of a cyclic group is cyclic. If |a| = n, then for each positive divisor k of n, a has exactly one subgroup of order k and no others. Theorem 7.1 Lagrange’s Theorem In a finite group the order of a subgroup divides the order of the group. Theorem 9.1 Normal Subgroup Test A subgroup H of G is normal in G if and only if xHx1 # H for all x in G. Theorem 10.3 First Isomorphism Theorem If f is a group homomorphism from G to a group, then G/Ker f  f(G). Theorem 11.1 Fundamental Theorem of Finite Abelian Groups Every finite Abelian group is a direct product of cyclic groups of prime-power order. Theorem 12.3 Subring Test A nonempty subset S of a ring R is a subring if a  b and ab are in S whenever a and b are in S. Theorem 13.4 Characteristic of an Integral Domain The characteristic of an integral domain is 0 or prime. Theorem 14.1 Ideal Test A nonempty subset A of a ring R is an ideal if a  b [ A whenever a and b are in A; and ra and ar are in A whenever a [ A and r [ R. Theorem 14.4 R/A is a Field if and only if A is Maximal Let R be a communitive ring with unity and let A be an ideal of R. Then R/A is a field if and only if A is maximal. Theorem 15.3 First Isomorphism Theorem for Rings If f is a ring homomorphism from R to a ring, then R/Ker f  f(R). Corollary 1 of Theorem 17.5 F[x]p(x) Is a Field Let F be a field and p(x) an irreducible polynomial over F. Then F[x]p(x) is a field. Theorem 21.5 [K : F]  [K : E][E : F] If K is a finite extension field of the field E and E is a finite extension field of the field F, then [K : F]  [K : E][E : F]. Theorem 22.2 Structure of Finite Fields The set of nonzero elements of a finite field is a cyclic group under multiplication. Theorem 24.3 Sylow’s First Theorem Let G be a finite group and p a prime. If pk divides |G|, then G has a subgroup of order pk. Theorem 24.5 Sylow’s Third Theorem The number of Sylow p-subgroups of G is equal to 1 modulo p and divides |G|. Furthermore, any two Sylow p-subgroups of G are conjugate.

Cayley Tables Cayley Table for the Dihedral Group of Order 6 R0 R120 R240 F F F

R0

R120

R240

F

F

F

R0 R120 R240 F F F

R120 R240 R0 F F F

R240 R0 R120 F F F

F F F R0 R120 R240

F F F R240 R0 R120

F F F R120 R240 R0

F'

F"

F Cayley Table for the Dihedral Group of Order 8 R0 R90 R180 R270 H V D D

R0

R90

R180

R270

H

V

D

D

R0 R90 R180 R270 H V D D

R90 R180 R270 R0 D D V H

R180 R270 R0 R90 V H D D

R270 R0 R90 R180 D D H V

H D V D R0 R180 R270 R90

V D H D R180 R0 R90 R270

D H D V R90 R270 R0 R180

D V D H R270 R90 R180 R0

D

V

D'

H

Notations (The number after the item indicates the page where the notation is defined.)

SET THEORY

SPECIAL SETS

>i[ISi
Z Q Q1 F* R R1 C

FUNCTIONS AND ARITHMETIC

ALGEBRAIC SYSTEMS

f 21 t|s tBs gcd(a, b) lcm(a, b) f(a) f: A → B gf, ab D4 Dn e Zn det A U(n)

Rn SL(2, F) GL(2, F) g21 2g UGU UgU H#G H,G kal Z(G)

intersection of sets Si , i [ I union of sets Si , i [ I {x [ S | x , a}, equivalence class of S containing a, 16 number of elements in the set of S integers, additive groups of integers, ring of integers rational numbers, field of rational numbers multiplicative group of positive rational numbers set of nonzero elements of F real numbers, field of real numbers multiplicative group of positive real numbers complex numbers the inverse of the function f t divides s, 3 t does not divide s, 3 greatest common divisor of the integers a and b, 5 least common multiple of the integers a and b, 6 image of a under f, 18 mapping of A to B, 18 composite function, 19 group of symmetries of a square, dihedral group of order 8, 31 dihedral group of order 2n, 32 identity element, 41 group {0, 1, . . . , n 2 1} under addition modulo n, 42 the determinant of A, 43 group of units modulo n (that is, the set of integers less than n and relatively prime to n under multiplication modulo n), 44 {(a1, a2, . . . , an) U a1, a2, . . . , an [ R}, 45 group of 2 3 2 matrices over F with determinant 1, 46 2 3 2 matrices of nonzero determinants with coefficients from the field F (the general linear group), 46 multiplicative inverse of g, 49 additive inverse of g, 49 order of the group G, 57 order of the element g, 57 subgroup inclusion, 58 subgroup H 2 G, 58 {an U n [ Z}, cyclic group generated by a, 61 {a [ G U ax 5 xa for all x in G}, the center of G, 62

C(a) C(H) f(n) N(H) cl(a) Gn Sn An G
{g [ G U ga 5 ag}, the centralizer of a in G, 64 {x [ G U xh 5 hx for all h [ H}, the centralizer of H, 66 Euler phi function of n, 79 {x [ G U xHx21 5 H} 5 {x [ G U Hx 5 xH}, the normalizer of H in G, 91 conjugacy class of a, 91 {gn U g [ G}, 92 group of one-to-one functions from {1, 2, ? ? ? , n} to itself, 97 alternating group of degree n, 106 G and G are isomorphic, 123 mapping given by fa(x) 5 axa21 for all x, 130 group of automorphisms of the group G, 131 group of inner automorphisms of G, 131 {ah U h [ H}, 138 {aha21 | h [ H}, 138 the index of H in G, 142 {f [ G U f(i) 5 i}, the stabilizer of i under the permutation group G, 145 {f(i) U f [ G}, the orbit of i under the permutation group G, 145 external direct product of groups G1, G2, . . . , Gn, 155 {x [ U(n) U x mod k 5 1}, 159 commutator subgroup, 174 H is a normal subgroup of G, 178 factor group, 180 {hk U h [ H, k [ K}, 188 internal direct product of H and K, 188 internal direct product of H1, . . . , Hn, 190 kernel of the homomorphism f, 200 inverse image of g9 under f, 202 inverse image of K under f, 203 ring of polynomials with integer coefficients, 238 ring of all 2 3 2 matrices with integer entries, 238 direct sum of rings, 239 ring of multiples of n, 241 ring of Gaussian integers, 241 group of units of the ring R, 243 ring of Gaussian integers modulo n, 245 characteristic of R, 252 principal ideal generated by a, 263 ideal generated by a1, a2, . . . , an, 263 factor ring, 263 sum of ideals A and B, 270 product of ideals A and B, 270 annihilator of A, 272 nil radical of A, 272 field of quotients of F[x], 286 ring of polynomials over R, 293

deg f (x) Fp(x) M2(Q) kv1, v2, . . . , vnl F(a1, a2, . . . , an) f 9(x) [E:F] GF( pn) GF( pn)* cl(a) Pr(G) np W(S) ka1, a2, . . . , an U w1 5 w2 5 . . . 5 wtl Q4 Q6 D` fix(f) Cay(S:G) k * (a, b, . . . , c) (n, k) Fn d(u, v) wt(u) Gal(E/F) EH Fn(x) C⊥

degree of the polynomial, 295 pth cyclotomic polynomial, 310 ring of 2 3 2 matrices over Q, 346 subspace spanned by v1, v2, . . . , vn, 347 extension of F by a1, a2, . . . , an, 357 the derivative of f (x), 362 degree of E over F, 372 Galois field of order pn, 383 nonzero elements of GF( pn), 384 {xax21 U x [ G}, the conjugacy class of a, 403 probability that two elements from G commute, 405 the number of Sylow p-subgroups of a group, 410 set of all words from S, 438 group with generators a1, a2, . . . , an and relations w1 5 w2 5 . . . 5 wt , 441 quarternions, 445 dicyclic group of order 12, 445 infinite dihedral group, 446 {i [ S U f(i) 5 i}, elements fixed by f, 489 Cayley digraph of the group G with generating set S, 498 concatenation of k copies of (a, b, . . . , c), 506 linear code, k-dimensional subspace of Fn, 523 F % F % ? ? ? % F, direct product of n copies of the field F, 523 Hamming distance between vectors u and v, 524 the number of nonzero components of the vector u (the Hamming weight of u), 524 the automorphism group of E fixing F, 546 fixed field of H, 546 nth cyclotomic polynomial, 562 dual code of a code C, 573